The smallest number is 4663.

Find the smallest number which when increased by 17 is exactly divisible

by both 520 and 468

Quick Scoop

We want a number xxx such that when you add 17 to it, the result is divisible by both 520 and 468.

That means:

  • x+17x+17x+17 must be a common multiple of 520 and 468.
  • To make x+17x+17x+17 as small as possible, we choose the least common multiple (LCM) of 520 and 468.

So:

  • LCM(520,468)=4680(520,468)=4680(520,468)=4680.
  • x+17=4680⇒x=4680−17=4663x+17=4680\Rightarrow x=4680-17=4663x+17=4680⇒x=4680−17=4663.

✅ Answer: 4663 is the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Step-by-step breakdown

1. Set up the condition

Let the required number be xxx.

We are told:

  • x+17x+17x+17 is divisible by 520.
  • x+17x+17x+17 is also divisible by 468.

So x+17x+17x+17 is a common multiple of 520 and 468.

Since we want the smallest such xxx, x+17x+17x+17 should be the least common multiple of 520 and 468.

2. Use LCM idea

From standard math practice on such questions:

  • LCM(520,468)=4680(520,468)=4680(520,468)=4680.

Now:

  • x+17=4680x+17=4680x+17=4680.
  • So x=4680−17=4663x=4680-17=4663x=4680−17=4663.

3. Quick check

You can verify:

  • 4663+17=46804663+17=46804663+17=4680.
  • 4680÷520=94680÷520=94680÷520=9 (an integer).
  • 4680÷468=104680÷468=104680÷468=10 (an integer).

So 4680 is divisible by both 520 and 468, and 4663 is exactly 17 less than that, matching the condition.

Mini facts table (for clarity)

html

<table>
  <tr>
    <th>Item</th>
    <th>Value</th>
  </tr>
  <tr>
    <td>Numbers given</td>
    <td>520 and 468 [web:3][web:5][web:7][web:9]</td>
  </tr>
  <tr>
    <td>LCM(520, 468)</td>
    <td>4680 [web:3][web:5][web:7][web:9]</td>
  </tr>
  <tr>
    <td>Condition</td>
    <td>x + 17 is divisible by 520 and 468 [web:3][web:9]</td>
  </tr>
  <tr>
    <td>Equation</td>
    <td>x + 17 = 4680 [web:3][web:9]</td>
  </tr>
  <tr>
    <td>Smallest number x</td>
    <td>4663 [web:3][web:5][web:7][web:9]</td>
  </tr>
</table>

TL;DR

  • Take the LCM of 520 and 468 → 4680.
  • Subtract 17 → 4680−17=46634680-17=46634680−17=4663.
  • So the required smallest number is 4663.

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