Question

Find the smallest number which when divided by 24,30,48,60 leave a remainder 2 in each xa

Answer 1

Hi there!
Here's the answer:

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¶ Type of problem:

• To find the smallest No. which when divided by x,y,z leaves a remainder 'r' in each case.


¶ Approach to problem:

Steps:

• Find LCM(x,y,z) (say it as 'l')
• Add it to Remainder 'r'

•°• Required No. = l + r

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Now,
Given 24, 30, 48, 60
& r = 2


• l = LCM(24,30,48,60)

3 | 24, 30, 48, 60
4 | 8, 10, 16, 20
2 | 2, 10, 4, 5
… | 1, 5, 2, 5

•°• l = 3×4×2×5×2×5 = 1200

•°• Required No. = 1200 + 2 = 1202

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¶ Remainder = divisor - [dividend × quotient]

Verification:

1202 - 24(50) = 2
1202 - 30(50) = 2
1202 - 48(50) = 2
1202 - 69(50) = 2


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:)

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Hope it helps