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31 Mar 2022, 09:39
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Difficulty:
95%
(hard)
Question Stats:
43% (02:20) correct
57%
(02:13)
wrong
based on 174
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Erin writes a list of every sixth integer in increasing order, starting with the positive integer x. Harry writes a list of every ninth integer in increasing order, starting with the positive integer y. Will any integer appear on both Erin’s list and Harry’s list?
1. x is a multiple of y.
2. x – y is a multiple of 3.
1. x is a multiple of y.
2. x – y is a multiple of 3.
17 Jun 2022, 05:55
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This question is actually a "trains catching-up" situation in disguise.
Train A starts from point X at the speed of 6 units per unit time.
Train B starts from point Y at the speed of 9 units per unit time.
Now, the question "will any integer appear in both the list?" is another way to ask whether the trains will catch up at any point. The constraint though is that the time after which the trains meet must be an integer. For eg, It can be 1h, 2m, or 3s. But it cannot be 1.25h, 2.5m, or 3.22s
Time after which the trains will meet = (difference in distance)/(difference in speed)
Or, steps after which the two lists will have a common integer = (difference between X and Y)/3
S1: Say X= Ny where N is an integer. => X-Y = (N-1)Y. So, in order to get a common integer in the two lists under this condition, either (N-1) or Y must be multiple of 3. For all other cases, the lists won't have a common integer.
S1 is insufficient.
S2: Difference between X and Y is multiple of 3. Sufficient.
Answer B
Train A starts from point X at the speed of 6 units per unit time.
Train B starts from point Y at the speed of 9 units per unit time.
Now, the question "will any integer appear in both the list?" is another way to ask whether the trains will catch up at any point. The constraint though is that the time after which the trains meet must be an integer. For eg, It can be 1h, 2m, or 3s. But it cannot be 1.25h, 2.5m, or 3.22s
Time after which the trains will meet = (difference in distance)/(difference in speed)
Or, steps after which the two lists will have a common integer = (difference between X and Y)/3
S1: Say X= Ny where N is an integer. => X-Y = (N-1)Y. So, in order to get a common integer in the two lists under this condition, either (N-1) or Y must be multiple of 3. For all other cases, the lists won't have a common integer.
S1 is insufficient.
S2: Difference between X and Y is multiple of 3. Sufficient.
Answer B
General Discussion
vaibhav1221
Joined: 19 Nov 2017
Last visit: 24 Jul 2025
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Location: India
Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
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WE:Account Management (Advertising and PR)
Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
Posts: 296
31 Mar 2022, 23:23
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Erin's sequence - x, x+6, x+12, x+18...
Harry's sequence - y, y+9, y+18, y+27...
Statement 1: x is a multiple of y.
The first thing that should strike you is that x ≥ y
CASE 1:
Assume x = y = 10
In this case the common numbers are 10, 28, 46 etc i.e. multiples of LCM of 9 and 6.
CASE 2:
Now, assume x = 20 and y = 10
There should be a difference of 10 between multiples of 9 and 6 for this to be true.
The unit's digit of a number that is a multiple of 6 could be 0,2,4,6,8 whereas the unit's digit of a number that is a multiple of 9 could be 0,1,2,3,4,5,6,7,8,9. The only time they meet is when the units digit is even and when that's the case multiple of 6 = multiple of 9. Eg: 18, 36, 54, 72, 90 etc. So, when x=20 and y=10, then there won't be any common number for the sequences.
Insufficient.
Statement 2: x – y is a multiple of 3.
Let's take a few examples
x=6, y=0 (x-y = 6)
6, 12, 18, 24
0, 9, 18, 27
x=13, y=10 (x-y = 3)
13, 19, 25
10, 19, 27
x=19, y=10 (x-y = 9)
19, 25, 31, 37
10, 19, 28, 37
x=10, y=10 (x-y = 0)
10, 16
10, 19
The pattern here is that if x-y = 3z, then (z+1)th term in both the sequences would be the same.
Because 9-6 is also 3 (common differences of both the sequences) and x-y is also a multiple of 3, you should be very confident that this statement must be true.
Sufficient.
Hence, B.
Harry's sequence - y, y+9, y+18, y+27...
Statement 1: x is a multiple of y.
The first thing that should strike you is that x ≥ y
CASE 1:
Assume x = y = 10
In this case the common numbers are 10, 28, 46 etc i.e. multiples of LCM of 9 and 6.
CASE 2:
Now, assume x = 20 and y = 10
There should be a difference of 10 between multiples of 9 and 6 for this to be true.
The unit's digit of a number that is a multiple of 6 could be 0,2,4,6,8 whereas the unit's digit of a number that is a multiple of 9 could be 0,1,2,3,4,5,6,7,8,9. The only time they meet is when the units digit is even and when that's the case multiple of 6 = multiple of 9. Eg: 18, 36, 54, 72, 90 etc. So, when x=20 and y=10, then there won't be any common number for the sequences.
Insufficient.
Statement 2: x – y is a multiple of 3.
Let's take a few examples
x=6, y=0 (x-y = 6)
6, 12, 18, 24
0, 9, 18, 27
x=13, y=10 (x-y = 3)
13, 19, 25
10, 19, 27
x=19, y=10 (x-y = 9)
19, 25, 31, 37
10, 19, 28, 37
x=10, y=10 (x-y = 0)
10, 16
10, 19
The pattern here is that if x-y = 3z, then (z+1)th term in both the sequences would be the same.
Because 9-6 is also 3 (common differences of both the sequences) and x-y is also a multiple of 3, you should be very confident that this statement must be true.
Sufficient.
Hence, B.
