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Difficult Math Problem #110 - Arithmetic

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800guy Master | Next Rank: 500 Posts
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Difficult Math Problem #110 - Arithmetic

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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rascal Newbie | Next Rank: 10 Posts
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29 Mar 2007
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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

RAGS Senior | Next Rank: 100 Posts
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Fri Mar 30, 2007 8:37 am
he is right

rascal Newbie | Next Rank: 10 Posts
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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

RAGS Senior | Next Rank: 100 Posts
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Fri Mar 30, 2007 8:37 am
he is right

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Roland2rule Master | Next Rank: 500 Posts
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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

Cybermusings Legendary Member
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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

800guy Master | Next Rank: 500 Posts
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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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