In the correctly worked addition problem shown, P, Q, R, and S are dig

3P5 =300+10P+5

03P5
+4QR
_____
08S4

Only value R can take is 9
.+1...+1
03.... P............ 5
+4.....Q........... 9
_______________
8.........S...........4

Hence P+Q+1 = 10+S

3P+1 = 10+S

3P+1-10= S

3(P-3) = S

Hence S must be multiple of 3; S can take only 1 value among all options, that is 3.

A

3P5
+4QR
8S4
5+r = 4 ;only possible when r = 9
given that Q=2P
and hundereds digit is 8 ; so we have +1 carry fwd from tens place
which means that P+Q+1= S+10
+1 on LHS is carry fwd from units place of 5+9 ; 14

3Q=S+9
S has to be 0,3,6,9
from given options S = 3 sufficient
OPTION A

3P5
+4QR
_____
8S4

We can easily deduce that the value of R will be 9, such that total becomes 5+9= 14
so, 1+P+Q= S [ 1, because of the carry from 14]
Given , Q=2P
so, 1+P+2P= S
=> S= 3P+1
We, see that the hundred's digit is 8, so we need to get a carry of 1 from the addition of 3P+1
Maximum value of P can be 4 and minimum value can be 3 [As Q=2P, and 1+Q+P>=10]
So, value of S can be either '0'(when, P=3) or '3' (when P=4)
from the options 3 fits the bill.
(Option A)

R must be 9 because the units digit of the sum, 5 + R, is 4.

3P5 + 4Q9 = 8S4

305 + 10P +409 + 10Q = 804 +10S

Since Q = 2P,

714 + 30P = 804 + 10S

30P - 90 = 10S

3P - 9 = S

S = 3(P - 3)

We can see that S must be divisible by 3, and there is only one answer choice divisible by 3.

Answer: A

My possible answer options are different with what is posted here.

A. 3
B. 4
C. 5
D. 7
E. 9

By looking at the options we can eliminate options D and E

from the question we can say

5 + R = 4, this happens of if R is 9

If R = 9, then there is a carry forward to the \(2^{nd}\) line.

in the third line we have 3 + 4 = 8, this happens only if there is a carry forward to the \(3^{nd}\) line from the \(2^{nd}\) line.

If there should be a carry forward from the second line then the value of P has to be either 3 or 4.

Since we know Q = 2P

If P = 3 then Q = 6 then P + Q + 1 = 3 + 6 + 1 = 10
so the possible value of S = 0

If P = 4 then Q = 8 then P + Q + 1 = 4 + 8 + 1 = 13
so the possible value of S = 3

Now the possible values of S are 0 or 3

According to the options the answer has to be Option A

Video solution by GMATinsight

I followed a very simple method:

S cannot be a number greater than 10 (i.e) it has to be between 0 and 9.

We have P. We have Q=2P. We know that P+Q=S. We also know that S<=9

Therefore,

P + Q <= 9
P+2P <= 9
3P <= 9
P <= 3

Therefore we have the following possibilities for P, Q, and S

P: 0 | 1 | 2 | 3
Q: 0 | 2 | 4 | 6
S: 0 | 3 | 6 | 9

3 is the only option in the answer choices. Answer is (A)

In unit digits, 5 + R = 4. Since 5 + R should be greater than 5, the only possibility would be 5 + R = 14 So there will be carry of 1 in tens place.
5 + R = 14
R = 9.

In Ten’s place, 1 + P + Q = S or S + 10 depending on whether there is a carry in hundred place or not.
In hundred place, 3 + 4 ≠ 8. That means there is carry of 1 in hundred’s place.
1 + 3 + 4 = 8

So, we can conclude that: 1 + P + Q = S + 10
Q = 2p
P + 2P = S + 10 -1
3P = S + 9
Since 3P is a multiple of 3, (S+9) also should be a multiple of 3. That means S should be a multiple of 3.
Only option A will satisfy that condition.


Hope this helps!
Clifin J Francis,
GMAT SME

Hey can someone please explain why we have done P+Q+1 = S+10
I understood the part P+Q+1=S but not why we have added 10 to the RHS.

How do we know r= positive number? As in options negative numbers are also given, why r cannot be equal to -1? Or does saying PQRS are digits (as in question) means only positive numbers?

Hello Vishlesh,


I understand your doubt and I appreciate the fact that you thought about it and brought it forward.

There is actually more than just this one question that I need to explain here. The GMAC rolled out two versions of this question – one is old (the one you posted on) and the other is new (from OG ‘21 onwards).

1. Old Question - on GMATClub
2. New Question – OG ‘21 onwards

Let’s begin with your question on the older version.

Old Question - on GMATClub

In this version, choices D and E have negative integers {-1 and -2}. These choices were especially given to induce the doubt that you also have in mind presently - “if S can be negative, R and other digits can probably be negative as well.”

But, as I said, it is just a means to create confusion and trap students who have conceptual gaps. Otherwise, think about it, have you ever seen a 3-digit number that looks like 8(-1)4 or 46(-1)? No, right?

By definition, digits are always non-negative. Specifically, when we say digits, we only mean the integers from 0 to 9, inclusive.

So, for this question in fact, choices D and E should be the ones rejected right away! If you check the choice selection data for this question (on GC - above the question), you will see that very few people got trapped this way. Choices D and E got only 1% of all votes each.


New Question – OG ’21 onwards

Now, in the Official Guides from 2021 and 2022, we have the same question BUT with different choices. This was most likely done to make the question more challenging since rejecting the negative choices was straightforward.
The new choices for the same question are:

1. 3
2. 4
3. 5
4. 7
5. 9

Try this one out! Tell me how it goes. 😊


Hope this helps!

Best Regards,
Ashish Arora,
Quant Expert, e-GMAT

Check Units Terms

5 + R = 4

Only choice is 5 + 9 = 14 ,

R = 4

1 is carried over

Check 100th Terms

3 + 4 = 8

Possible only if 1 is carried over from P + Q addition Tenth Terms.

10th Terms

1 (carried over from Units addition) + P + Q > = 10

P + Q > = 9

P + 2P > = 9

3 P > = 9

P > = 3

If P > = 3, possible values of P are 9, 8, 7, 6, 5, 4, 3

If so Qs corresponding possible values are 18, 16, 14, 12, 10, 8, 6

But Q cannot be 2 Digit and has to be single digit

So P & Q can be 4 and 8 or 3 and 6.

If P = 4, Q = 8, then 1+P+Q = 13, hence S = 3

If P = 3, Q = 6, then 1+P+Q = 10, hence S = 0

Only 3 is available in the option.

So answer is 3.