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What is the smallest five-digit number which when divided by 7, 11 and 12 Jan 2022, 09:45
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E
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35% (medium)

Question Stats:

74% (02:34) correct 26% (02:10) wrong based on 92 sessions

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What is the smallest five-digit number which when divided by 7, 11 and 21 leaves a remainder of 5 in each case?

A. 10159
B. 10008
C. 10109
D. 10097
E. 10169

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E
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What is the smallest five-digit number which when divided by 7, 11 and 12 Jan 2022, 23:57
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There's a problem in this question.
The answer choices should be in ascending or descending order.

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AnuragNair
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What is the smallest five-digit number which when divided by 7, 11 and 13 Jan 2022, 00:39
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LCM of 7,11,21 = 231

Smallest five digit number = 10000

Nearest multiple of 231 = 10164

Since the number leaves a remainder of five, the number should be = 10164+5 = 10169

Option E.
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What is the smallest five-digit number which when divided by 7, 11 and 26 Apr 2024, 23:56
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



What is the smallest five-digit number which when divided by 7, 11 and 21 leaves a remainder of 5 in each case?

Let's solve the problem using two methods

Method 1: Elimination

Let's subtract 5 from all the option choices so that we have numbers which are divisible by 7, 11 and 21

=> A. = 10154 (10159-4)
=> B. = 10003
=> C. = 10104
=> D. = 10092
=> E. = 10164

Since the numbers are divisible by 21 => they are divisible by 3
=> Sum of the digits should be divisible by 3
(Watch this video to MASTER DIVISIBILITY RULES)

=> A. Sum of digits = 1+0+1+5+4  = 11 NOT DIVISIBLE by 3 => ELIMINATE
=> B. Sum of digits = 1+0+0+0+3 = 4 NOT DIVISIBLE by 3 => ELIMINATE
=> C. Sum of digits = 1+0+1+0+4 = 6 DIVISIBLE by 3 => KEEP
=> D. Sum of digits = 1+0+0+9+2 = 12 DIVISIBLE by 3 => KEEP
=> E. Sum of digits = 1+0+1+6+4 = 12 IVISIBLE by 3 => KEEP

Now, numbers are divisbile by 11
=> 9999 + 99 = 10098 is divisible by 11

=> C. = 10104 = 10098 + 6 = NOT DIVISIBLE by 11 => ELIMINATE
=> D. = 10092 = 10098 - 6 = NOT DIVISIBLE by 11 => ELIMINATE
=> E. = 10164 = ANSWER

Method 2: Logic

Since the number leaves a remainder of 5 when divided by 7, 11, 21
=> Number will be 5 + a amultiple of LCM(7,11,21)
= 5 + 11*21k = 5 + 231k
( Watch this video to MASTER LCM and GCD)

Now, a multiple of 231 closer to 10000 = 231*44 = 10164

And 10164 + 5 = 10169

So, Answer will be E
Hope it helps!

Watch the following video to MASTER Remainders
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