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Determine the sum of all single-digit replacements for n such that the number {42{,}789{,}n37} is divisible by 3

User ReynierPM
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2 Answers

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To determine the sum of all single-digit replacements for n such that the number {42,789,n37} is divisible by 3, you need to find the values of n that make the sum of the digits divisible by 3. Here are the steps:

First, find the sum of the digits in the number {42,789,n37} without replacing n.

4 + 2 + 7 + 8 + 9 + n + 3 + 7 = 30 + n

Now, you want the sum of the digits to be divisible by 3, so you need to find the possible values of n that make 30 + n divisible by 3.

30 + n ≡ 0 (mod 3)

To make 30 + n divisible by 3, n must be equal to -30 modulo 3.

n ≡ -30 (mod 3)

Find the remainder when -30 is divided by 3:

-30 ÷ 3 = -10

So, n ≡ -10 (mod 3).

Now, you need to find all single-digit replacements for n that satisfy this congruence:

n ≡ -10 (mod 3)

The single-digit numbers that satisfy this congruence are:

n = 1 (because -10 + 1 = -9, which is divisible by 3)

n = 4 (because -10 + 4 = -6, which is divisible by 3)

n = 7 (because -10 + 7 = -3, which is divisible by 3)

Now, calculate the sum of these possible values for n:

1 + 4 + 7 = 12

So, the sum of all single-digit replacements for n such that the number {42,789,n37} is divisible by 3 is 12.

User Nidhi
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2 votes

Answer:

1275563

is prime, but 1+2+7+5+5+6+3=29

which is different from 1,2,4,5,7,8

. Did you mean that we add the digits on and on until the sum is a one digit number ?

Explanation:

User Stav Bodik
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7.7k points