Chapter 16 – Playing With Numbers

• In case of XY, the original number is 10X + Y, and upon reversal, it becomes 10Y + X. When both numbers are added, the result will always be divisible by 11, and the remainder will be zero.
  Sum of the both the numbers = (10Y + X) + (10X + Y)
  = 10X + Y + 10Y + X
  = 11X +11Y
  = 11(X + Y)
  For example, Adding 85 and 58, we get 143 which is divisible by 11 and the remainder is zero.
• When both the original and reversed numbers are subtracted, the result is always divisible by 9. For example, 85 – 58 =27, which is divisible by nine and the remainder is zero.

• In case of XYZ, the original number is 100X +10Y + Z, and upon reversal, the numbers become 100Z +10Y + X. 
  Difference of the both the numbers = (100X +10Y + Z) – (100Z +10Y + X)
  = 100X + 10Y + Z – 100Z – 10Y – X
  = 99X – 0 – 99Z
  = 99(X – Z)
  Consequently, the answer will always be a multiple of 99 and divisible by 11. 

• Think of any 3-digit number and form two more different numbers from the same digits. Add these numbers and divide by 37. You will get zero remainder. 
• Let us take PQR, then the other numbers would be QRP (the unit’s place digit moved to the left end), and RPQ (the hundred’s digit moved to the right end).
  PQR = 100P + 10Q + R
  QRP = 100Q + 10R + P
  RPQ = 100R + 10P + Q
         = 111 (P + Q + R) 
         = 37 × 3 (P + Q + R), divisible by 37.
• For example, 839 + 398 + 983 = 2220/37 = 60, 0 remainder.

Letters for Digits

These puzzles are like cracking a code because we substitute some letters for digits. You have to solve the problem to find out which letter represents which digit, keeping some essential rules in mind.

• One letter can represent only one digit.
• The first digit cannot be ZERO, i.e., you cannot write XY as 00XY or 0XY.

Example 1- Puzzles with Addition
Let us find the value of X in 2X8 + 73X = 982

• In the above addition, 8 + X = 2, that means X = 4. 
• Now put 4 in place of X in the equation, 248 + 734 = 982.

Example 2- Puzzles with Multiplication

Let us find the value of X and Y in XY x X7 = 129Y

• Y × 7 = Y, it can be only possible is Y = 0 or Y = 5
• Now for X, if we assume that X = 1 then the maximum value of XY x X7 would be 19 × 17 = 323. Therefore, the value of X cannot be 1. Similarly, If X = 2, then the maximum value of XY x X7 would be less. Therefore, the value of X cannot be 2.
• If X = 3 And Y= 0, XY × X7 = 30 × 37 = 1110.
• If X = 3 And Y = 5, XY × X7 = 35 × 37 = 1295. 
• The second possibility yields the correct answer; therefore, X = 3 and Y = 5.