A stamp collector has the habit to arrange or rearrange the stamps accordingly. while doing this he some times keeps the stamps in pairs, or in group of 3 or in 4 or in or in 6 and realises that in any case he is left with 1 stamp and when he arranges them in groups of 7 no stamps remain. what is the number of stamps he has?
answer can also b 121, it is the least no. of stamps possible........
if one stamp is left if he makes groups of 2,3,4,6 then we have to find LCM of 2,3,4,6 which is 12 now since we get a remainder of 1 ... it shud b 12+1 ie 13
now we now tht the no. of stamps is exactly divisible by 7 therefore no. of stamps = 13*7=121
RE: A stamp collector has the habit to arrange or rear...
He keeps stamps in pairs 3's 4's 5's and in 6's then 1 is left. when he aranges them in 7's then nothing is left so the number of stamps must be multiple of 7 so (2*3*4*5*6) 720 therefore number of stamps are (720+1) 721 which is divisible by 7.
RE: A stamp collector has the habit to arrange or rear...
721 is correct but least possible no. is 427.49 is wrong b'coz It has to give 1 as remainder when divided by 5. (not printed in the problem).Correct me if I'm wrong.....
RE: A stamp collector has the habit to arrange or rear...
the answer should be a multiple of 7 bt not of 2 3 4 or 6 and the number should always be one more than yhe multiples of 2 3 4and 6... so 7 -ruled out. 14 ruled out 21- ruled out.so on ... 49 is right... (2*24+1) (3*16+1) (4*12+1) (6*8+1).. so v can conclude that the lest possible number is 49.
RE: A stamp collector has the habit to arrange or rear...
There can multiple answers for the question.... Mathematically (n-1) LCM(2 3 4 5 6)*k1 > (n-1) 60*k1 ---->(1) and n 7*k2 ----->(2) solve (1) & (2) we have n 30*k1 + (7*k2+1)/2 where k1 (7*k2-1) 60few solutions:with k1 5 and k2 43 n 301 ... least possible solution... with k1 12 and k2 103n 721and etc....Noticible: number k2 has 3 as last digit.
RE: A stamp collector has the habit to arrange or rear...
answer can also b 121 it is the least no. of stamps possible........
if one stamp is left if he makes groups of 2 3 4 6 then we have to find LCM of 2 3 4 6 which is 12 now since we get a remainder of 1 ... it shud b 12+1 ie 13
now we now tht the no. of stamps is exactly divisible by 7 therefore no. of stamps 13*7 121
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