An eight digit number 15785xy3 is divisible by 9, then find the maximu

1.	An eight digit number 15785xy3 is divisible by 9, then find the maximum possible value of 2x + 3y.1. 412. 243. 404. 43
Answer» Correct Answer - Option 1 : 41 Given: The number 15785xy3 is divisible by 9. Concept used: A number is divisible by 9 when sum of it’s digits is divisible by 9. Calculation: Sum of all digits = 1 + 5 + 7 + 8 + 5 + x + y + 3 ⇒ Sum of digits = 29 + x + y As the number is divisible by 9, so 29 + x + y is divisible by 9 Multiple of 9 nearest to 29 is 36 29 + x +y = 36 ⇒ x + y = 36 – 29 = 7 But this is not the maximum value of x and y Next nearest multiple of 9 is 45 29 + x +y = 45 ⇒ x + y = 16 For 2x + 3y to be maximum y should be greater than x For y = 9, x = 16 – 9 = 7 2x + 3y = 2 × 7 + 3 × 9 = 41 ∴ Maximum value of 2x + 3y is 41

An eight digit number 15785xy3 is divisible by 9, then find the maximum possible value of 2x + 3y.1. 412. 243. 404. 43

Answer» Correct Answer - Option 1 : 41

Given:

The number 15785xy3 is divisible by 9.

Concept used:

A number is divisible by 9 when sum of it’s digits is divisible by 9.

Calculation:

Sum of all digits = 1 + 5 + 7 + 8 + 5 + x + y + 3

⇒ Sum of digits = 29 + x + y

As the number is divisible by 9, so 29 + x + y is divisible by 9

Multiple of 9 nearest to 29 is 36

29 + x +y = 36

⇒ x + y = 36 – 29 = 7

But this is not the maximum value of x and y

Next nearest multiple of 9 is 45

29 + x +y = 45

⇒ x + y = 16

For 2x + 3y to be maximum y should be greater than x

For y = 9, x = 16 – 9 = 7

2x + 3y = 2 × 7 + 3 × 9 = 41

∴ Maximum value of 2x + 3y is 41

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