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​

Answer 1

Answer: sum of A and B = 7

the no. that is divisible by 7,11,13 completely =437437

here, A=3 and B= 4

so, A+B =3+4 =7

therefore , correct ans...7.

Step-by-step explanation:

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Answer 2

Answer:

Here we will find the sum of the numbers of A and B using given details.

Final Answer is: A + B = 7

Step-by-step explanation:

• A six digit number 4A7B3C is divisible by $7,11,13$ with the remainder $0$.
• First we should find the L.C.M of $7,11,13$. L.C.M of $7,11,13$ is $1001$.
• $7 * 11* 13 = 1001$, when any 3 digit number multiplied by $1001$ then repeats itself and always completely divisible by $7,11,13$.
• Let we consider the 3 digit number be $abc$.

                     $a b c * 1001 = 4A7B3C$

                           $a b c a b c = 4A7B3C$

Now compare on both sides,   $a = 4, c = 7, b = 3$

Hence that six digit number is $a b c a b c = 437437$

                       $4A7B3C = 437437$

by comparing both sides, $A = 3, B = 4, C = 7$

Now we can find the sum of numbers of A, B​.

That is,               $A + B = 3+4$

                          $A+ B = 7$

Final Answer: sum of numbers of A, B: A+ B = 7