Question

A six digit number 4A7B3C is such that it leaves remainder 0 when divided by 7, 11 and 13. What will be the sum of numbers at A, B and C?​

Answer 1

The sum of numbers at A, B and C is 14

Given :

A six digit number 4A7B3C is such that it leaves remainder 0 when divided by 7, 11 and 13.

To find :

The sum of numbers at A, B and C

Solution :

Step 1 of 2 :

Write down the given data

Here it is given that a six digit number 4A7B3C is such that it leaves remainder 0 when divided by 7, 11 and 13.

LCM of 7, 11 and 13 = 1001

Since the six digit number 4A7B3C is divisible by 7, 11 and 13

∴ The six digit number 4A7B3C is divisible by 1001

Step 2 of 2 :

Find the sum of numbers at A, B and C

The six digit number 4A7B3C is divisible by 1001

We know that when a three digit is multiplied by 1001 then it repeats it iself

Let xyz be the three digit number

∴ xyz × 1001 = 4A7B3C

⇒ xyzxyz = 4A7B3C

Comparing both sides we get

x = 4 , y = 3 , z = 7

∴ A = y = 3

B = x = 4

C = z = 7

Hence the required sum

= A + B + C

= 3 + 4 + 7

= 14

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