A number theory problem by Paola Ramírez

Number Theory Level 2

Let be $n$ a positive integer which satisfies the following conditions:

$n$ is even.
$n$ divided by 5 leaves a remainder of 1.
$n$ is divisible by 7.
$n<1000$ .
The sum of digits of $n$ is 23.

Find $n$ .

The answer is 896.

1 solution

Guilherme Niedu
May 13, 2016

Lets call the digits of $n$ (n has up to three digits, since $n<1000$ ) as $a, b, c$ (i.e. $n = \overline{abc}$ ). If $n$ has a remainder of $1$ when divided by $5$ , $c$ is either $1$ or $6$ .

Since $n$ is even, $\fbox{c = 6}$

The digits sum up to $23$ , which means that $a + b + c = 23$ , or $a + b = 17$ . Since $a$ and $b$ are digits, they are natural numbers between $0$ and $9$ , which means that either $a = 8$ and $b = 9$ ( $n = 896$ ) or $a = 9$ and $b = 8$ ( $n = 986$ ).

$986$ is not divisible by $7$ , so the answer is $\fbox{896}$

n is congruent to 0 mod 7 and 1 mod 5. Solve simultaneous linear congruences using Chinese remainder theorem (as 7 and 5 are co prime integers) to get a n is congruent to 21 mod 5. Keep adding 35 to the remainder to find all possibilities. 896 is only possible solution for n

Elliot Ede - 2 years ago

A number theory problem by Paola Ramírez

The answer is 896.

1 solution

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