I don't get the explanation that is given for the following question, can someone explain in their own terms...
If a and b are both single-digit positive integers, is a + b a multiple of 3?
(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.
(2) a – 2b is a multiple of 3.
OA is D...PLEASE HELP!!!!
MGMAT - CAT #4, Q#20
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stmt 1 - the number ab is a multiple of 3...straightfwd..if a number has to be a multiple of 3 then the sum of digits has to be a multiple of 3...so a+b is a multiple of 3.......Suff
stmt 2 - a-2b is a multiple of 3. now if a+b is a multiple of 3 then either the sum or the difference or the product HAS to be a multiple of 3.You can verify with examples...here we'll take the difference of 2 numbers. Difference = (a-2b) - (a+b) = -3b....definitely a multiple of 3 and so a+b has to be a multiple of 3....Suff
IMO ans is D
stmt 2 - a-2b is a multiple of 3. now if a+b is a multiple of 3 then either the sum or the difference or the product HAS to be a multiple of 3.You can verify with examples...here we'll take the difference of 2 numbers. Difference = (a-2b) - (a+b) = -3b....definitely a multiple of 3 and so a+b has to be a multiple of 3....Suff
IMO ans is D
Trying hard!!!
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Hi Nisha,
I dont know whether this is the best way, but I would solve it like this
(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.
now there is a mathematical rule that "A nos is divisible by 3 if the sum of its digits is divisible by 3" (this is true always)
i.e a+b is divisible by 3
SUFF
(2) a – 2b is a multiple of 3.
when taking multiples & fators we only use natural nos i.e 1 & above
hence a-2b cannor be -ve
also a - 2b = a + b - 3b
i.e (a+b) - 3b
here 3b is divisible by 3 i.e it is a multiplr of 3
now since it is substacted from (a+b) hence a+b must be a multiple of 3
i.e 3* somenos - 3*somenos will be a multiple of 3
henece SUFF
D
I dont know whether this is the best way, but I would solve it like this
(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.
now there is a mathematical rule that "A nos is divisible by 3 if the sum of its digits is divisible by 3" (this is true always)
i.e a+b is divisible by 3
SUFF
(2) a – 2b is a multiple of 3.
when taking multiples & fators we only use natural nos i.e 1 & above
hence a-2b cannor be -ve
also a - 2b = a + b - 3b
i.e (a+b) - 3b
here 3b is divisible by 3 i.e it is a multiplr of 3
now since it is substacted from (a+b) hence a+b must be a multiple of 3
i.e 3* somenos - 3*somenos will be a multiple of 3
henece SUFF
D
Regards
Samir
Samir
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Can we do it this way:-
(a-2b+3b-3b)/3 = k( adding & subtractoing 3b)
( (a+b) - 3b ) /3 = k
(a+b) / 3 - b = k
(a+b) /3 = b + k (since b is given as single digit integer, so b + k will be another integer, say n)
(a+b)/3 = n, hence a+b is divisible by 3.
This is clear for explanations provided earlier.1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.
(a-2b)/3 = k (k is any integer)(2) a – 2b is a multiple of 3.
(a-2b+3b-3b)/3 = k( adding & subtractoing 3b)
( (a+b) - 3b ) /3 = k
(a+b) / 3 - b = k
(a+b) /3 = b + k (since b is given as single digit integer, so b + k will be another integer, say n)
(a+b)/3 = n, hence a+b is divisible by 3.