Question

A number when divided by 11 and 12 gives 7 and 8 as remainders respectively, find the

number.

Answer 1

Let the number be x. Given that when x is divided by 7, 8 & 9 the remainder is 1, 2 & 3 respectively. But if you add 6 to that number, then it would be divisible by all three of them (7, 8 & 9)
Since,
1 + 6 = 7
2 + 6 = 8
3 + 6 = 9
But we are dealing with smallest such number. So, x + 6 would be the 'least common multiple (1cm) of 7, 8 & 9 which is 7*8*9 = 504. x + 504 which implies x = 498.
Therefore, smallest number which does the job is 498. Similar, such numbers are 2*504 - 6 = 1002, 3*504 - 6 = 1506, 4*504 - 6 = 2010 and so on.

Answer 2

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Let the number is x.

According to question,

When x divided by 11 it leaves 7 as remainder.

Assume that when x is divided by 11 then quotient is a.

So,

 x = 11a + 7   ---------------- ( 1 ).

Now,

When x is divided by 12 then it leaves 8 as remainder.

Assume that when x is divided by 12 then quotient is b.

So,

  x = 12b + 8  ------- ( 2 )

From ( 1 ) and ( 2 ) , we get,

⇒ 11a + 7 = 12b + 8

⇒ 11a - 12b = 8 - 7

⇒ 11a - 12b = 1

∴  11a = 1 + 12b.

Here we have to find such values of a and b such that they are natural numbers.

When , b = 10.

⇒11a=1 + 12b

⇒11a = 1 + 12 × 10

⇒ 11a = 1 + 120

⇒ 11a = 121

⇒ a = 121 ÷ 11

∴ a = 11.

By substituting the value of a in ( 1 ),

⇒ x = 11a + 7

⇒ x = 11 × 11 + 7

⇒ x = 121 + 7

∴  x = 128

By substituting the value of b in ( 2 ),

⇒ x = 12b + 8

⇒ x = 12 × 10 + 8

⇒ x = 120 + 8

∴ x = 128.

The required number is 128.