7 Additions and 1 Subtraction

Fill the boxes below with only 7 addition (+) and 1 subtraction (-) operators. 

The smartest way to find the right place 

Correct Placement of Subtraction Operator

This was the question!

For a moment. let's put '+' operator in all given boxes.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 33.

Let 'x' be the number ahead of which '-' should be there where x can be any number from 1 to 9. 

In above addition the x is already (unknown value) added so first we need to subtract it twice from LHS; first to compensate the addition operation & second for the mandatory subtraction operator that is to be placed at right place.

So for example, if it's ahead of 4 as right place then

1 + 2 + 3 + 4 - 4 - 4 + 5 + 6 + 7 + 8 + 9 = 33.

1 + 2 + 3 + (4 - 4) - 4 + 5 + 6 + 7 + 8 + 9 = 33.

For x, which is already there & can be any from 1 to 9,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9  - x - x = 33. 

45 - 2x = 33

2x = 12

x = 6.

Hence, the subtraction (-) sign should be ahead of 6. 

Crack The 10-Digits Password

In an attempt to further protect his secret chicken recipe, Colonel Sanders has locked it away in a safe with a 10-digit password.  Unable to resist his crispy fried chicken, you'd like to crack the code. 

Try your luck using the following information:

All digits from 0 to 9 are used exactly once.

No digit is the same as the position which it occupies in the sequence.

The sum of the 5th and 10th digits is a square number other than 9.

The sum of the 9th and 10th digits is a cube.

The 2nd digit is an even number.

Zero is in the 3rd position.

The sum of the 4th, 6th and 8th digits is a single digit number.

The sum of the 2nd and 5th digits is a triangle number.

Number 8 is 2 spaces away from the number 9 (one in between).

The number 3 is next to the 9 but not the zero.

Click here to know how to CRACK it! 

Cracking The 10-Digits Password

What was the challenge? 

Let's suppose the 10-digits password is ABCDEFGHIJ.

---------------------------------

Take a look at the given clues to crack down the password.

1. All digits from 0 to 9 are used exactly once.

2. No digit is the same as the position which it occupies in the sequence.

3. The sum of the 5th and 10th digits is a square number other than 9.
 
4. The sum of the 9th and 10th digits is a cube.
 
5. The 2nd digit is an even number.
 
6. Zero is in the 3rd position.
 
7. The sum of the 4th, 6th and 8th digits is a single digit number.
 
8. The sum of the 2nd and 5th digits is a triangle number.
 
9. Number 8 is 2 spaces away from the number 9 (one in between).
 
10. The number 3 is next to the 9 but not the zero. 

--------------------------------------

STEPS : 

Respective Clues are in ().

1] Since, every digit is used only once (1), the sum of 2 digits can't exceed 
9 + 8 = 17 and minimum sum of 2 digits can be only 0 + 1 = 1.

2] So, the sum of 5th & 10th digits which is square (3) can be - 1, 4, 16.
The sum of 9th & 10th digits which is cube (4) can be - 1, 8.
The sum of 2nd & 5th digits which is triangle number (8) can be - 1, 3, 6, 10, 15.

3] As per (6), 0 is already at third position i.e. C = 0. Hence, 2nd, 5th, 9th or 10th digits can't be 0. Therefore, the sum of 5th & 10th or 9th & 10th or 2nd & 5th can't be 1. Revising list of possible values of those in step 2.

The sum of 5th & 10th digits which is square (3) can be -  4, 16.
The sum of 9th & 10th digits which is cube (4) can be -  8.
The sum of 2nd & 5th digits which is triangle number (8) can be -  3, 6, 10, 15.

4] Possible pairs of 9th & 10th digits - (1,7), (7,1), (2,6), (6,2), (3,5), (5,3)

If IJ = 71, then 5th digit E must be 4 - 1 = 3 & possible values of 2nd digit B are 0,3,7.
But as per (5), B must be even & C = 0 already.

IJ = 26 or 62 or 35 doesn't leave any valid value for 5th digit E.

If IJ = 53, then 5th digit E must be 1 & possible values of 2nd digit B are 2,5,9
That is B = 2 (5). But as per (2), no digit is the same as the position which it occupies in the sequence. So, the number 2 can't be at 2nd position where B is there.

Hence, IJ must be 17 i.e. I = 1, J = 7.

5] This leaves only 9 as valid value for 5th digit E so that (3) is true. 

So, E = 9 and hence B = 6 with sum of 2nd digit B and 5th digit E as 15.

6] So far, we have, B = 6, C = 0, E = 9, I = 1 and J = 7. 

As per (10), the number 3 is next to the 9 but not the zero. Hence, the 6th digit F must be 3. F = 3.

7] As per (9), the 8 is 2 spaces away from 9 i.e. here 5th digit E=9. With 3rd position already occupied by C = 0, the digit 8 must be at 7th position. 
So, G = 8. 

8] As per (7), D + F + H must be single digit. We have, F = 3 already, 
so possible values of D + H are 0, 1, 2, 3, 4, 5, 6.

D + H can't be 0 (0+0), 1 (1+0), 2(1+1/2+0), 3(1+2/3+0), 4(2+2/3+1/4+0), 5(2+3/4+1/5+0) since digits are repeated with C=0, I=1, F=3 already.

Hence, D + H = 6. 

9] Now, D = 5 and H = 1 is not possible. Also, D/H can't be 0 or 6. Both D and H can't be 3 at the same time. D can't be 4 as D is at 4th position. 

Hence, D = 2, H = 4.

10] With B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7, the only digit left for the first position A is 5. So A = 5.

CONCLUSION : 

A = 5, B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7.

So, the 10-digits password ABCDEFGHIJ is 5602938417.

 

The Numbered Neighbors - Puzzle

AJ, Celeste, Juan, Kara, Lily, and Randy all live on Redwood Avenue. Each of their house numbers has three digits, but the only digits in their house numbers are 2, 3, 5, and 6. The same digit may appear twice in one of the addresses.
AJ's house number is Juan's house number doubled. 

Juan lives next to Lily and right across from Randy. 

Randy's number is the lowest on the street. 

Celeste's number is a higher number than AJ's but has the same three digits as his. They both have the same digit in the hundreds place. 

Kara's house number is the reverse of Juan's. 

What is each person's street address?

View Answer! 

The Numbered Neighbors Puzzle - Solution

What was the puzzle?

As we know, AJ, Celeste, Juan, Kara, Lily, and Randy all live on Redwood Avenue. Each of their house numbers has three digits, but the only digits in their house numbers are 2, 3, 5, and 6 with repetition of digits in house number allowed. 

Let's have all the given clues once again here.

1. AJ's house number is Juan's house number doubled. 

2. Juan lives next to Lily and right across from Randy. 

3. Randy's number is the lowest on the street. 

4. Celeste's number is a higher number than AJ's but has the same three digits as his. They both have the same digit in the hundreds place.
 
5. Kara's house number is the reverse of Juan's.

STEPS : 

1] As per (1), AJ's house number is double of Juan's house number. From the given sets of digits i.e. 2, 3, 5, 6, this is only possible if there is 3 at third place of Juan's house number and 6 at third place of AJ's house number.

That is AJ's house number is XX6 and that of Juan's is XX3.

2] As we can see carry is not being forwarded from Juan's house number XX3 while getting doubled to give AJ's house number. So, it's clear that, the number formed by first two digits of AJ's house number is exactly double of the number formed by first two digits of Juan's house number.

From the given set of digits 2, 3, 5, 6, possible house numbers of AJ and Juan are 666 & 333 or 526 & 263.

If AJ is having house number 666, then as per (4), there would be no valid house number left for Celeste using digits 2, 3, 5, 6.

Therefore, AJ's house number must be 526 and that of Juan's must be 263

3] Now, as per (2), Randy must be having house number 262.

4] As per (4), Celeste's house number must be 5XX. Since, his house number has same digits as that of AJ's but higher than AJ's house number, he must be having house number 562.

5] As per (5), Kara's house number is reverse of Juan's. Therefore, Kara's house number is 362.

6] And as per (2), for the fact that Juan is living next to the Lily to be true, Lily must have house number 265 using the given set of digits 2, 3, 5, 6.

CONCLUSION :

The street addresses of all are listed as below - 

AJ         : 526, Redwood Avenue.
 
Celeste : 562, Redwood Avenue.
 
Juan     : 263, Redwood Avenue.
 
Kara     : 362, Redwood Avenue.
 
Lily       : 265, Redwood Avenue.
 
Randy   : 262, Redwood Avenue.

 

Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and the first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

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1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


------------------------------------------------------- 

STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.

  

The Case of Fourth Mystery Number

There is a ten-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. 

Given the following clues, what is the number?

1) B + C + F + J = E + G + H + I = AD
2) B - H = J - G = 3
3) C - F = E - I = 5
4) B * I = AJ

Here are steps demystifying the mystery number! 

The First Case of Mystery Number 

The Second Case of Mystery Number 

The Third Case of Mystery Number

Demystefying The Fourth Mystery Number

What was the challenge? 

Given are hints to identify number ABCDEFGHIJ.

---------------------------------------------------- 

1) B + C + F + J = E + G + H + I = AD

2) B - H = J - G = 3

3) C - F = E - I = 5

4) B * I = AJ

---------------------------------------------------- 

STEPS : 

---------------------------------------------------- 


STEP 1 :

The sum of digits from 0 to 9 is 45.

Maximum value of AD = 98 and 

Minimum value of AD = 10 (A can't be 0 as leading 0's not allowed).

If AD = 98 then sum of rest of digits B + C + F + J + E + G + H + I must be 
45 - (9 + 8) = 28.

If AD = 10 then sum of rest of digits B + C + F + J + E + G + H + I must be 
45 - (1 + 0) = 44.

The sum of such 8 digits is divided into 2 parts in form of
B + C + F + J  and E + G + H + I which in turn must be equal to AD.

Therefore, each group of four digits must sum to one of the following: 14, 15, 16, 17, 18, 19, 20, 21, 22 (with AD varying from 28 to 44).

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STEP 2 :

Using Trial And Error method to get possible values of AD.

If AD = 14 then B + C + F + J + E + G + H + I = 45 - (1 + 4) = 40 and

B + C + F + J  = E + G + H + I = 40/2 = 20 = AD but AD = 14 assumed.

Hence, this value of AD is invalid.

Similarly, 15, 16, 17, 19, 20, 22 are invalid values of AD leaving behind only 18 and 21 as possible values.

Hence, A must be either 1 or 2 and D must be either 1 or 8.

That is either A or D takes 1.

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STEP 3 : 

As per Hint 2, B and J > H and G by 3 respectively and since 1 already taken by A or D,

Possible Values of B and J - 3, 5, 6, 7, 8, 9.

Possible Values of H and G - 2, 3, 4, 5, 6.

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STEP 4 : 

As per Hint 3, C and E > F and I by 3 respectively and since 1 already taken by A or D,

Possible Values of C and E - 5, 7, 8, 9.

Possible Values of F and I -  0, 2, 3, 4.

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STEP 5 : 

So with B having possible values as 3, 5, 6, 7, 8, 9 and I having possible values as 0, 2, 3, 4 the equation B * I has following 24 possibilities - 

(3 x 0 = 0) (3 x 2 = 6) (3 x 3 = 9) (3 x 4 = 12)

(5 x 0 = 0) (5 x 2 = 10) (5 x 3 = 15) (5 x 4 = 20) 
  
(6 x 0 = 0) (6 x 2 = 12) (6 x 3 = 18) (6 x 4 = 24)

(7 x 0 = 0) (7 x 2 = 14) (7 x 3 = 21) (7 x 4 = 28) 

(8 x 0 = 0) (8 x 2 = 16) (8 x 3 = 24) (8 x 4 = 32) 

(9 x 0 = 0) (9 x 2 = 18) (9 x 3 = 27) (9 x 4 = 36).

B * I can't be 0 as AJ can't be 0. Also, since A can't be 0 the product B*I can't be single digit like 6 or 9. 

Moreover, A has to be either 1 or 2 as deduced in STEP 2 and J must be among 3, 5, 6, 7, 8, 9 as deduced in STEP 3. 

Now the equation B * I = AJ has possibilities as - 

(5 x 3 = 15) (6 x 3 = 18)
  
(7 x 4 = 28) (8 x 2 = 16) 

(9 x 2 = 18) (9 x 3 = 27)  

Since (5 x 3 = 15) suggests that B = J = 5 which is against the rule that no 2 alphabets can take same digit. Hence, that possibility is eliminated.

Revised Possible Values of B - 6, 7, 8, 9.

Revised Possible Values of I - 2, 3, 4.

Revised Possible Values of J - 6, 7, 8. 

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STEP 6 :  

Since I can't be 0, E can't be 5 as E - I = 5.

Revised Possible Values of E - 7, 8, 9.

Since J - G = 3, and if J is among 6, 7, 8

Revised Possible Values of G - 3, 4, 5.

Since B - H = 3, and if B is among 6, 7, 8, 9

Revised Possible Values of H - 3, 4, 5, 6.

So letters A, B, D, E, G, H, I and J together takes digits 1, 2, 3, 4, 6, 7, 8 and 9 not in order though.

This leaves behind only possible value of C = 5 and F = 0. 

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STEP 7 :

Now, since H can't be 5 hence B can't be 8. Also, G too can't be 5 so J can't be 8 too. So both B and J can't be 8. 

Now, revising B * I = AJ possibilities deduced in STEP 4 as - 

 (9 x 3 = 27)  

Leaves only possible valid combination thereby.

So, we get, B = 9, I = 3, A = 2 and J = 7.

---------------------------------------------------- 

STEP 8 : 

If B = 9 then H = 6.

If I = 3 then E = 8.

If J = 7 then G = 4.

The equation B + C + F + J = 9 + 5 + 0 + 7 = 21 = AD gives A = 2 and D = 1.

---------------------------------------------------- 

CONCLUSION :

A = 2, B = 9, C = 5, D = 1, E = 8, F = 0, G = 4, H = 6, I = 3, J = 7.

Hence, the mystery number ABCDEFGHIJ is 2951804637.

 Verifying the given hints - 

1) B + C + F + J = E + G + H + I = AD     
    9 + 5 + 0 + 7 = 8 + 4 + 6 + 3 = 21

2) B - H = J - G = 3
    8 - 3 = 5 - 0 = 5 

3) C - F = E - I = 5
    7 - 4 = 9 - 6 = 3 

4) B * I = AJ 
    9 * 3 = 27