↧↧↧ 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: EliminationLet'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 =>
KEEPNow, 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 =
ANSWERMethod 2: LogicSince 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 EHope it helps!
Watch the following video to MASTER Remainders