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Difficult Math Problem #110 - Arithmetic

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Difficult Math Problem #110 - Arithmetic

by 800guy » Fri Mar 16, 2007 9:00 am
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

A. 36
B. 28
C. 12
D. 9
E. None

from diff math doc. answer coming when a some people respond with explanations
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by Neo2000 » Fri Mar 16, 2007 9:57 am
Ideally, if the two numbers are added and then divided then the remainder should be the same as if they were divided individually and then added.

However, this gives us a remainder of 4 only, which means that from 11+21 = 32 only 4 remains => that 28 was completely divided by the divisor.

Therefore 28 was the divisor??
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by gabriel » Fri Mar 16, 2007 10:15 am
... yup .... 28 it is ...
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OA

by 800guy » Mon Mar 19, 2007 9:43 am
OA:

Let the divisor be a.

x = a*n + 11 ---- (1)
y = a*m + 21 ----- (2)
also given, (x+y) = a*p + 4 ------ (3)
adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4)

equate 3 and 4.
a*p + 4 = a*(n+m) + 32
or
a*p + 4 = [a*(n+m) + 28] + 4
cancel 4 on both sides.
u will end up with.
a*p = a*(n+m) + 28.

which implies that 28 should be divisible by a. or in short a = 28 works.
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by Cybermusings » Tue Mar 27, 2007 10:10 am
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

A. 36
B. 28
C. 12
D. 9
E. None

Here we can easily eliminate Choice C and Choice D...Choice D because the divisor is less than both remainders (i.e. 11 and 21)...That's not possible

Likewise 21 as a remainder is less than 12...Hence eliminate D

Let us take 28...When the remainder of two numbers divided by a common divisor are 11 and 21 respectively then when we add these two numbers and divide it by the divisor the remainder should be the sum of the earlier 2 remainders...

11+21 = 32. Which when divided by 28 leaves a remainder 4. Hence I would go with B...I am not too good at explaining but I am pretty sure its the right answer
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easily solved -gone in 60 seconds

by rascal » Thu Mar 29, 2007 5:20 pm
this is simple..
step 1) add 11 and 21
step 2 ) Subtract 4
Step 3 ) get the answer - 11 + 21 = 32
32 - 4 = 28
Step 4 ) Mark the answer ...

Cheers
Rascal
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by RAGS » Fri Mar 30, 2007 8:37 am
he is right
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by BTGmoderatorRO » Fri Sep 29, 2017 3:12 pm
Let the different number be P and Q.
Let the divisor be represented by A.
let us take some real life assumption in solving solving this question. The assumptions would be stated as we progress.
Recall from the question:
1) P/A gives reminder 11
2) Q/A gives remainder 21
3) (P+Q)/A gives remainder 4

Let digress a bit by considering this short example below:
say P=15 & A=2
15/2= 7 r 1 (r=remainder)
finding an expression for 15 in terms of the quotient (7), remainder(1) and the divisor (2).

Now going back to the question,
P/A= 1 r 11
NB: we assume that A can only go in P once.
P=11+ (A*1)
P=A+11 ------------(i)

similarly with the same assumption,
Q/A=1r21
Q=21+(A*1)
Q=A+21------------(ii)

from i and ii, we can get equation representing the value of A.
A=P-11 ---------iii
A=Q-21 ----------iv

since A=A, the P-11=Q-21 -------------v
Note that from the question, it was given that (P+Q)/A gives r=4
so, from eqn (v), P-Q = 11-21
P=Q=11+21 (assumption stated)
P+Q=32

32/A= 1 r4
assuming that A can be found once in 32.
from the example that 32=4+(1*A)
32=4+A
A=28

Thus, the divisor is 28. option B is the correct option
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