The First Case of 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) A + B + C + D + E is a multiple of 6.

2) F + G + H + I + J is a multiple of 5.

3) A + C + E + G + I is a multiple of 9.

4) B + D + F + H + J is a multiple of 2.

5) AB is a multiple of 3.

6) CD is a multiple of 4.

7) EF is a multiple of 7.

8) GH is a multiple of 8.

9) IJ is a multiple of 10.

10) FE, HC, and JA are all prime numbers.

NOTE : AB, CD, EF, GH and IJ are the numbers having 2 digits and not product of 2 digits like A and B, C and D .....

HERE is that MYSTERY number! 

Demystifying The First Mystery Number

What was the challenge?

Take a look at the clues given for identifying the number ABCDEFGHIJ.

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1) A + B + C + D + E is a multiple of 6.
 
2) F + G + H + I + J is a multiple of 5.
 
3) A + C + E + G + I is a multiple of 9.
 
4) B + D + F + H + J is a multiple of 2.
 
5) AB is a multiple of 3.
 
6) CD is a multiple of 4.
 
7) EF is a multiple of 7.
 
8) GH is a multiple of 8.
 
9) IJ is a multiple of 10.
 
10) FE, HC, and JA are all prime numbers.

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STEPS :  

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STEP 1 : Since, the digits in number ABCDEFGHIJ are from 0 to 9 with no repeat, the sum of all digits must be 0 + 1 + .....+ 9 = 45.

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STEP 2 : In first 2 conditions, it's clear that all digits of mystery number are added i.e. from A to J. However, addition of first 5 digits is multiple of 6 and addition of rest of digits is multiple of 5. 

That means the total addition of 45 must be divided into 2 parts such that one is multiple of 6 & other is multiple of 5.

30 and 45 is only pair that can satisfy these conditions. Hence,

A + B + C + D + E = 30.

F + G + H + I + J = 15.

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STEP 3 : In next 2 conditions, sums of digits at odd positions and even positions are listed. Moreover, the sum of digits at odd positions has to be multiple of 9 & that of at even positions need to be multiple of 2.

So again,  the total addition of 45 must be divided into 2 parts such that one is multiple of 9 & other is multiple of 2.

The only pair to get these conditions true is 27 and 18. Hence, 

A + C + E + G + I = 27.

B + D + F + H + J = 18.

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STEP 4 : As per condition 9, IJ is multiple of 10. For that, J has to be 0 and with that now 0 can't be anywhere else. J = 0. 

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STEP 5 : Since, one digit can be used only once, numbers like 11, 22, 33....are eliminated straightaway.

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STEP 6 : As per condition 10, JA is prime number. With J = 0, for JA to be prime number, A = 2, 3, 5, 7. 
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STEP 7 : As per condition 5, AB is a multiple of 3. 

Let's list out possible value of AB without any 0, possible digits of A = 2, 3, 5, 7 and excluding numbers having 2 same digits as -

  21, 24, 27, 36, 39, 51, 54, 57, 72, 75, 78. 

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STEP 8 : For numbers FE and HC to be prime (as per condition 10), C and E can't be 0, 5 or even.

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STEP 9 : As per condition 6, CD is multiple of 4 and as per condition 8, GH is multiple of 8. So, D and H has to be even digits.

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STEP 10 : As per condition 6, CD is a multiple of 4. So the possible values of CD without 0, with C not equal to 5 and with odd C, even D -

  12, 16, 32, 36, 72, 76, 92, 96. 

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STEP 11 : As deduced in STEP 3 , B + D + F + H + J = 18.  

With J = 0 and D, H as even digits (STEP 9), both B and F has to odd or even to get to the even total of 18. 

If both of them are even then the total of 

B + D + F + H + J  = 2 + 4 + 6 + 8 + 0 = 20.

which is against our deduction.

Hence, B and F must be odd numbers. 

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STEP 12 : So, possible values of AB deduced in STEP 7 are revised with odd B as -

  21, 27, 39, 51, 57, 75.

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STEP 13 : As per condition 7, EF is a multiple of 7. With F as odd (STEP 11), along with E as odd, not equal to 5 (STEP - 8), possible value of EF are - 

  21, 49, 63, 91.

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STEP 14 : But as per condition 10, FE is PRIME number. Hence, the only possible value of EF from above step is 91. SO, E = 9 and F = 1.

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STEP 15 : Now after 1 and 9 already taken by F and E, possible value of AB in STEP 12 are again revised as - 27, 57, 75. And it's clear that either A or B takes digit 7. So 7 can't be used further.

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STEP 16 : So after 7 taken by A or B, E = 9, F = 1 possible values of CD deduced in STEP 10 are revised as - 32, 36.  Hence, C = 3.

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STEP 17 : With AB = 27, CD can't be 32. And if AB = 27, CD = 36 then,

A + B + C + D + E = 2 + 7 + 3 + 6 + D + 1 = 30.

D = 13.

This value of D is impossible.

Moreover, if CD = 32 and AB = 75 or 57, 

A + B + C + D + E = 5 + 7 + 3 + 2 + D + 1 = 30.

D = 12.

Again, this value of D is invalid. 

Hence, CD = 36 i.e. C = 3 and D = 6 and AB = 57 or 75 but not 27.

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STEP 18 : With AB = 57 or 75, CD = 36, EF = 91, J = 0, possible values of GH which is multiple of 8 (condition 8) are -  24, 48. 

That means either G or H takes 4. Or G is either 2 or 4.

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STEP 19 :  Now as deduced in STEP 3,

A + C + E + G + I = 27. 

A + 3 + 9 + G + I = 27

A + G + I = 15.
 
The letter G must be either 2 or 4 and A may be 5 or 7.

If A = 5, G = 4 then I = 6

If A = 7, G = 2 then I = 6

But we have D = 6 already, hence both of above are invalid.

If A = 7, G = 4 then I = 4.

Again, this is invalid as 2 letters G and I taking same digit 4.

Hence, A = 5, G = 2 is only valid combination thereby giving I = 8.

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STEP 20 : 

If A = 5, then B = 7 ( STEP 17 ). 

C = 3, D = 6 ( STEP 17 ).

E = 9, F = 1 ( STEP 14).

If G = 2, then H = 4 ( STEP 18 & 19).

I = 8 (STEP 19), J = 0 ( STEP 4). 

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CONCLUSION :

Hence, the mystery number ABCDEFGHI is 5736912480.

In the end, just to verify if the number that we have deduced is following all given conditions, 

1) 5 + 7 + 3 + 6 + 9 = 30 is  a multiple of 6.
2) 1 + 2 + 4 + 8 + 0 = 15 is a multiple of 5.
3) 5 + 3 + 9 + 2 + 8 = 27 is a multiple of 9.
4) 7 + 6 + 1 + 4 + 0 = 18 is a multiple of 2.
5) 57 is a multiple of 3.
6) 36 is a multiple of 4.
7) 91 is a multiple of 7.
8) 24 is a multiple of 8.
9) 80 is a multiple of 10.
10) 19, 43, and 05 are prime numbers.

Find Correct Digits For Correct Letters

If each letter represents a different nonzero digit, what must Z be?

Finding difficult? Click Here! 

Correct Digits For Correct Letters

The equation in question was....

First of all X + Y + Z  = Z < = 10 in not possible since in that case, X +  Y + Z - Z =0 i.e. 
X + Y = 0.

Hence, there must be carry 1 forwarded to digit's place. So,

X + Y + Z = 10 + Z

X + Y = 10.          ........(1)

Therefore, possible pairs for (X,Y) or for (Y,X) are (1,9), (2,8), (3,7), (4,6) and (5,5) out of which (5,5) is invalid as repeating digits are not allowed.

Now, since carry is forwarded to ten's place addition, it looks like,

1 + X + Y + Z = XY

Putting (1) in above,

1 + 10 + Z = XY

11 + Z = XY        ........(2)

The maximum value of Z can be 9 & in that case as per above equation XY = 20. But since 0 is not allowed Z must be less than or equal to 8. Then XY <= 11 + 8 = 19.

So X which seems to be carry to hundred place must be 1 (can't be 0 or 2 as proved above). Hence, X = 1.

If X = 1, then from (1), Y = 9.

And if XY = 19 then from (2) Z = 8.

To conclude, X = 1, Y = 9 and Z = 8.

Final equation, looks like,

11 + 99 + 88 = 198.

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Another Method :

The given equation is nothing but

10X + X + 10Y + Y + 10Z + Z = 100X + 10Y + Z

11(X + Y + Z) = 100X + 10Y + Z

89X = Y + 10Z = 10Z + Y

Since, only whole digits are allowed, maximum value of 10Z + Y can be 98 with Z = 9 & Y = 8. So if X = 2 then 89 x 2 = 198 = 10Z + Y is impossible case. Hence, X = 1.

And only Z = 8 and Y = 9 satisfies 89 x 1 = 10Z + Y = 10(8) + 9 = 89.

In short, X = 1, Y = 9 and Z = 8.

What Could Be The Product?

Zach chooses five numbers from the set {1, 2, 3, 4, 5, 6, 7} and tells their product to Claudia. She finds that this is not enough information to tell whether the sum of Zach’s numbers is even or odd. What is the product that Zach tells Claudia?

Here is the way to get that product!

Guessing The Correct Product in Question!

What was the question?

When Zach tells the product of 5 numbers that he has chosen he indirectly conveying product of 2 un chosen numbers.

For example, product of all numbers in set = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 and if Zach tell product 1 x 2 x 4 x 5 x 7 = 280 then obviously the product of numbers that he hasn't chosen is 5040/280 = 18 = 3 x 6.

Now, there are only 2 products viz. 12 and 6 which have more than 1 pair of numbers. 

The product 12 can be from pairs - (3,4) and (6,2)

The product 6 can be from pairs - (1,6) and (2,3) 

Here if the sum of un chosen numbers is odd (or even) then sum of other 5 chosen numbers also must be odd (or even).

In above cases, 6 has pairs whose sum is odd always and hence sum of other 5 numbers would be odd. In that case, Claudia would have been sure with if sum of numbers selected by Zach is either odd or even.

While in other case, the product 12 has 1 pair having sum odd (3,4) and other pair having sum even (6,2). Hence, Zach must have 'indirectly' suggested product 12 as a product of un chosen numbers that's why Claudia is saying that she doesn't know if the sum of numbers selected by Zach is even or odd.

Hence, the product of numbers selected by Zach = 5040 / 12 = 420.    

 

Formations of Special 6-Digit Numbers

How many six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place?

Skip to the count!

Counting The Formations of 6-Digit Special Numbers

You may need to read the question first! 

Let's remind that the numbers can't be repeated. So mathematically there are total 6! (720) unique numbers can be formed.

The number XXXXX1 will be always divisible by 1; so there we have 5! = 120 numbers.

The number XXXXX2 will be always divisible by 2; so there we have 5! = 120 numbers. 

Since sum of all digits is 21 which is divisible by 3; the number XXXXX3 will be always divisible by 3. So we have 5! = 120 more such numbers.

The number XXXXY4 is divisible only when Y = 2 or 6. So in the case we have 2 x 4! = 48 numbers.

The number XXXXX5 will be always divisible by 5; so there we have 5! = 120 numbers.

The number XXXXX6 will be always divisible by 6 (since it is divisible by 2 & 3); so there we have 5! = 120 numbers.

Adding all the above counts - 120 + 120 + 120 + 48 + 120 + 120 = 648.  

So there are 648 six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place.
  

One More CryptArithmatic Problem

What three digits are represented by X, Y, and Z in this addition problem?

  XZY

+XYZ
______
  YZX

Here is the solution!

Solution Of One More CryptArithmatic Problem

What was the problem?

Here is the equation rewritten.

  XZY
+XYZ
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  YZX
 

Let's start with the tens place. Z + Y = Z is there. That means either Y = 0 or 9 if 1 is carry from ones place.

Since Y is at hundred's place it can't be 0. Hence, Y = 9.

At hundred's column, now we have, X + X = 9. That's only possible if X = 4 and carry 1 forwarded from tens place. So X = 4.

Now, finally, at ones place, we have, 9 + Z = 4. Hence, Z must be 5 with carry 1 being forwarded to the next place.

To sum up, X = 4, Y = 9 and Z = 5.

 

Numbers For The Letters?

A - B = B

B * C = A

D : B = E

C * C = E

C + E = A

For which numbers stand A, B, C, D and E? 

Skip To The Answer!  

Correct Numbers For Letters

What was the question? 

Let's first rewrite all equations and number them

1. A - B = B

2. B * C = A

3. D : B = E

4. C * C = E

5. C + E = A

For which numbers stand A, B, C, D and E? 

From 1, A = 2B. Putting this in 2 gives,

C = 2.

Putting C = 2 in 4 gives, E = 4.

So from 5, A = C + E = 2 + 4 = 6.

Equation 2 gives, B = A/C = 6/2 = 3.

And equation 3 gives D = B * E = 3 * 4 = 12.

Conclusion : A = 6, B = 3, C = 2, D = 12 and E = 4.