Find the Smallest Number Which When Increased by 17 Is Divisible by 520 and 468

Video Explanation

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Solution

Question: Find the smallest number that, when increased by 17, is exactly divisible by both 520 and 468.

Step 1: Identify the Mathematical Concept

If a number, when increased by 17, is divisible by both 520 and 468, then the increased number must be a common multiple of 520 and 468.

The smallest such number is their LCM (Least Common Multiple).

Step 2: Find the LCM of 520 and 468

Prime factorisation:

520 = 2³ × 5 × 13

468 = 2² × 3² × 13

LCM = 2³ × 3² × 5 × 13

LCM = 4680

Step 3: Find the Required Smallest Number

Required number = 4680 − 17

Required number = 4663

Final Answer

∴ The smallest number which, when increased by 17, is exactly divisible by both 520 and 468 is 4663.

Conclusion

Thus, by finding the LCM of 520 and 468 and subtracting 17, we obtain the required smallest number as 4663.