When a number is divided by 15 20 and 35 each time the remainder is 8 then that smallest number is?

Answered by Guest on 2018-01-20 03:45:37 | Votes 5 | #

right answer is 428

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Answered by Guest on 2018-04-09 19:24:26 | Votes 2 | #

Using Rule 5, LCM of 15, 20 and 35 = 420 \ Required least number = 420 + 8 = 428

Explanation:

Given numbers

15, 20 and 35

The smallest number will be the LCM (Least Common Multiple) of these three numbers.

Let us find the prime factors for each number

15 = 3 × 5

20 = 2 × 2 × 5

35 = 5 × 7

The LCM of the three numbers is given by

LCM = 2 × 2 × 5 × 3 × 7 = 420

420 is exactly divisible by 15, 20, and 35.

Since we have a remainder of 8 in each case, the smallest number in each case will be given by

= 420 + 8 = 428.

When a number is divided by 15, 20 or 35, each time the remainder is 8. Then the smallest number is

673 5f080953429a37679ec10d09

1
328
false
2
338
false
3
428
true
4
427
false

when-a-number-is-divided-by-15-20-or-35-each-time-the-remainder-is-8-tSingleChoice5f080953429a37679ec10d094

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Answer : 3. "428 "

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Home » Aptitude » LCM and HCF » Question

When a number is divided by 15, 20 or 35, each time the remainder is 8. Then the smallest number is

As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .LCM of 15, 20 and 35 ( k ) = 420Here , remainder = 8

∴ Required least number = 420 + 8 = 428