A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
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Solution:
The number is 58.
LCM of 3, 4, 5 and 6 is 60. So these numbers when divides 60 gives remainder 0. Now to get the remainder 1, 2, 3 and 4 respectively, the number should be 2 (3-1 ; 4-2 ; 5-3 ; 6-4 ) short of 60.
The number is 58.
LCM of 3, 4, 5 and 6 is 60. So these numbers when divides 60 gives remainder 0. Now to get the remainder 1, 2, 3 and 4 respectively, the number should be 2 (3-1 ; 4-2 ; 5-3 ; 6-4 ) short of 60.
58 % 3 = 1
ReplyDelete58 % 4 = 2
58 % 5 = 3
58 % 6 = 4
so answer 58
lcm of 3,4,5,6 is 60
ReplyDeleteall multiples(60)-2 is the answer i.e
60n-2
58,118,178........ are d answers
ans is 58
ReplyDelete58 is the answer.Just do calculation.List out the number i.e x%6=4.
ReplyDelete@ rajuy4cs209
ReplyDeletecan u pls explain how did u knew to subtract 2 from 60n
@rahul..
ReplyDeleteThe ans given by @raju4cs209 is excellent.
1)LCM
2)ans=LCM-prev no. 0f small no(i.e )
if the same qustion for 4,5,6.Then ans=60-3
Ya raju's ans. is unique.... but @ shiba, why take the prev no. to the smallest? i mean whats the logic
ReplyDeleteits 58 or any multiple of 58
ReplyDelete58,118,178.......
ReplyDeleteThis comment has been removed by the author.
ReplyDelete@ shiba n rohmen
ReplyDeleteit is not d previous no to d small no
it is d difference b/w d remainder n its dividend
here it is 2
dis rule is for problems in which d difference b/w remainder n its dividend is constant
diff b/w 3 n 1 is 2
diff b/w 4 n 2 is 2
diff b/w 5 n 3 is 2
diff b/w 6 n 4 is 2
ans is lcm(nos)n-diff ie 60n-2