Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination. 

What is the maximum sum of money you could accumulate before the drawing ends?

THIS should be the maximum sum! 

Source

The Maximum That You Can Get!

What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.

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.

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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. 

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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 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.

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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

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STEPS : 

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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.

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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.

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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