Question

If the number 2345 a 60b is exactly divisible by 3 and 5 then the maximum value a+b

Answer 1

Hey there!!

If we want it divisible by 5 , the last digit will have to be 0 or 5

so, b = 0 , 5

Divisible by 3 = add all numbers and divide it by 3

2 + 3 + 4 + 5 + a + 6 + 0 + 0/5

Let us take 0

We get 20 + a

If we take a = 1 , then it will be 21 which is exactly divisible by 3

Now, a + b = 0 + 1 = ' 1 '

If we take b = 5

Then , the sum would be 25 + a

a = 2 , then 27 , divisible by 3

Then, a + b = 2 + 5 = ' 7 '

Two answers , 1 and 7

Least = 1

Required answer = 1

please mark as brainliest answer