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a and b are positive integers. What is the remainder when b

Expert replies
by Max@Math Revolution » Thu Jul 25, 2019 12:11 am

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A

B

C

D

E

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[GMAT math practice question]

a and b are positive integers. What is the remainder when b is divided by 4?

1) if a is divided by 4, the remainder is 3
2) if a^2+b is divided by 4, the remainder is 1
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Source: — Data Sufficiency |

by deloitte247 » Sat Jul 27, 2019 12:33 pm
$$a=b\%\ of\ c=\frac{b}{10}\cdot c$$
what is c ?
Statement 1
a is c percent of b
a is c% of b
$$\frac{c}{100}of\ b$$
value of c is still unknown
statement 1 is INSUFFICIENT.

Statement 2
b is c percent of a
$$b\ is\ c\%\ of\ a$$
$$\frac{c}{100}of\ a\ $$
$$a=\frac{b}{100}\cdot c$$
$$b=\frac{c}{100}\cdot\frac{b}{100}\cdot c$$
$$b=\frac{bc^2}{100^2}$$
$$\frac{100^2b}{b}=\frac{bc^2}{b}$$
$$\sqrt{c^2}=\sqrt{100^2}$$
$$c=100$$
Statement 2 alone is SUFFICIENT
$$answer\ is\ Option\ B$$
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by deloitte247 » Sat Jul 27, 2019 5:03 pm
Statement 1
If a is divided by 4 , the remainder is 3.

If a = 3 then 1/4, the remainder is 3.
If a = 7 then 7/4, the remainder is 3.
If a = 11 then 11/4, the remainder is 3.
Information given is not ENOUGH to arrive at a specified answer. hence statement 1 is INSUFFICIENT.

Statement 2
$$If\ a^2+b\ ids\ divided\ by\ 4,\ the\ remainder\ is\ 1$$
If a = 1 and b = 4
$$\frac{\left(1^2+4\right)}{4}=\frac{5}{4,\ remainder\ 1}$$b =
If a = 2 and b = 1
$$\frac{\left(2^2+1\right)}{4}=\frac{5}{4}$$
If a = 3 and b=4 $$\frac{\left(3^2+4\right)}{4}=\frac{13}{4}\ remainder\ is\ 1$$

Possible value of b is either 4 or 1 but no definite answer for the value of b ; statement 2 is NOT SUFFICIENT.

Combining both statement together
From statement 1 ; a is an odd number .
Both statements are SUFFICIENT.

$$Answer\ is\ Option\ C$$
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by Max@Math Revolution » Sun Jul 28, 2019 4:55 pm
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (a and b) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1) tells us that a has remainder 3, when it is divided by 4. So, a^2 has remainder 1, when it is divided by 4.
Condition 2) tells us that a^2+b has remainder 1 when it is divided by 4. Since a^2 has remainder 1 when it is divided by 4, b has remainder 0 when it is divided by 4.
Thus, both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since there is no information about b in condition 1), it is not sufficient.

Condition 2)
If a = 1 and b = 4, then b has remainder 0 when it is divided by 4.
If a = 4 and b = 1, then b has remainder 1 when it is divided by 4.
Condition 2) is not sufficient since it does not yield a unique answer.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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