Notes for Playing with Numbers

A 2 digit number can always be written as a combination of 2 different numbers.

for eg:–

65 = 10 × 6 + 5 → 6 is at tens place and 5 is at ones place.

23 = 10 × 2 + 3 → 2 is at tens place and 3 is at ones place

Thus, any two digit number can be written in a general form as 10 × x + y

Similarly,

572 = 5 × 100 + 7 × 10 + 2

123 = 1 × 100 + 2 × 10 + 3

The three digit number xyz can be written as 100 × x + 10 × y + z

Puzzles and games are a source of entertainment and education that makes interesting and challenging situations.

Step-1 : Choose any 2-digit number of the form 10 x + y.

Step-2 : Reverse the digits to get a new number i.e.,

Step-3: Add the reversed number to the original number.

(10x + y) + (10y +x) = 11x + 11y = 11(x +y)

Step-4 : Divide the answer by 11.

11(x + y) ÷ 11 = (x + y)

Adding both the number, we get 36 + 63 = 99, which is exactly divisible by 11

Step-1 : Choose a two digit number in the form 10x + y.

Step-2 : Reverse the digits to get a new number in the form 10y + x.

Step-3 : Subtract both the numbers.

(10y + x) – (10x + y) = 9y – 9x = (9 (y – x)

Step-4 : Divide the answer by 9.

9(y – x) ÷ 9 = (y – x)

Step-1 : Choose any three-digit number xyz in the form 100x + 10y + z

Step-2 : From 2 more numbers in a way

yzx = 100z + 10x + y.

Step-3 : Add all the three numbers

(100x + 10y + z) + (100y + 10z + x)

+ (100z + 10x + y)

Step-4 : Divide the answer by 111.

= 111 (x + y + z) ÷ 111 = (x + y + z).

Step-1 : Take any three-digit number xyz in the form 100x + 10y + z.

Step-2 : Reverse the digits : zyx = 100z + 10y + x.

Step-3 : Substract both the numbers.

(100x + 10y + z) – (100z + 10y + x)

= 99x – 99z = 99(x – z)

Every game has same rules. So, there are some rules for such puzzles also. There are two rules for solving them.

(i) Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.

(ii) The first digit of a number cannot be zero.

Solve for Q:

From the addition above, we can say Q + 3 = 1. For this, Q must be equal to 8. So, the puzzle becomes:

Find the digits A, B and C.

Since the one�s digit of B × 3 is B, it must be B = 0 or B = 5.

Now, for A

If A = 1

These two are not possible because C cannot be zero.

If A = 2

These two are not possible because C cannot be zero.

If A = 3

If is not possible because A is not zero and C cannot be zero.

If A = 5

A number is divisible by 2, if its unit digit is 0 or divisible by 2 i.e., 2, 4, 6, 8

Find the condition when a two-digit number xy and a 3-digit number xyz will be exactly divisibly by 2.

2-digit number xy can be written as 10x + y. 2 will always divide 10x.

So, 10x + y will be exactly divisible by 2 if y = 0, 2, 4, 6 or 8.

A 3-digit number xyz can be written as 100x + 10y + z. We can say, 2 will always divide 100x and 10y. So,

100x + 10y + z will be divisible by 2 if z = 0, 2, 4, 6 or 8.

A number is divisible by 3, if the sum of its digits is divisible by 3.

A number is divisible by 5, if the digit in its unit�s place is 5 or zero.

A number is divisible by 9, if the sum of its digits is divisible by 9.

If number is divisible by 10, if the digit at unit’s place is zero.