If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?
1) yz = 3
2) x is odd.
Plz help me solve it....
1) yz = 3
2) x is odd.
Plz help me solve it....
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remainder when 7xyz is divided by 4?
Source: Beat The GMAT — Data Sufficiency |
stat 1) from this statement we can just understand that this No. is not divisible by 4 but about the remainder we can not determine and the remainder can be 1,2, or 3 ==> insufficient
stat 2) knowing about X alone can not give us any clue abt remainder.. insufficient
together: we can make different No such as 7313,7331, 7513,7531 and the like... we can not determine the remainder bcuz it is 1 or 3
Ans: E
stat 2) knowing about X alone can not give us any clue abt remainder.. insufficient
together: we can make different No such as 7313,7331, 7513,7531 and the like... we can not determine the remainder bcuz it is 1 or 3
Ans: E
I think 7xyz is not a 4 digit no rather it is 7*x*y*z. Reason being no where it is mentioned that x,y,z are single digit integers and also if we consider them as single digit numbers then Stmt 1 will go wrong.
Well, you know that it is the wrong answer and yet you don't have the right answer.. interesting 
Let me see if I understood this problem correctly
Statement 1
N=7x3. If we substitute different values for x, I'll keep it simple and pick single digits for x
We get different remainders. For 703, the remainder is 3. For 713, the remainder is 1. Insufficient
Statement 2 is insufficient as we could pick just about any integers for yz. Insufficient
Statement 1 and 2
N=713, the remainder is 1, for N=733, the remainder is 1, for N=7133, the remainder is 1. It looks good. Sufficient
Let me see if I understood this problem correctly
Statement 1
N=7x3. If we substitute different values for x, I'll keep it simple and pick single digits for x
We get different remainders. For 703, the remainder is 3. For 713, the remainder is 1. Insufficient
Statement 2 is insufficient as we could pick just about any integers for yz. Insufficient
Statement 1 and 2
N=713, the remainder is 1, for N=733, the remainder is 1, for N=7133, the remainder is 1. It looks good. Sufficient
sl750 wrote:Well, you know that it is the wrong answer and yet you don't have the right answer.. interesting
Let me see if I understood this problem correctly
Statement 1
N=7x3. If we substitute different values for x, I'll keep it simple and pick single digits for x
We get different remainders. For 703, the remainder is 3. For 713, the remainder is 1. Insufficient
Statement 2 is insufficient as we could pick just about any integers for yz. Insufficient
Statement 1 and 2
N=713, the remainder is 1, for N=733, the remainder is 1, for N=7133, the remainder is 1. It looks good. Sufficient
what a bout 753? the remainder is 0.. then two statements together are still insufficient
on the other hand, i think the Q asks about 4-digit No.s
for 753 also remainder is 1 for all odd value of x remainder is 1 so C
If my post helped you- let me know by pushing the thanks button. Thanks
What is the remainder when 7xyz is divided by 4?
(1)yz=3. Putting this value in 7xyz we have
21x where x could be any value eg. 1,2,3,4 etc
When x is 1 21/4 gives a remainder of 1
When x is 2 42/4 gives a remainder of 2
When x is 3 63/4 gives a remainder of 3 etc
No fixed answer INSUFFICIENT
(2)x is odd
7x=ODD for sure as both are odd
yz could be odd or even
If both are odd then 7xyz is odd so remainder could be 1,2,3
If both are even then 7xyz is even(as Even*Even=Even)
And if 7xyz is even then remainder could be 0 or 2 INSUFFICIENT
Combining both we have
21x and that x is odd. So 21x is odd i.e. 7xyz is odd
x could be 1,3,5,7 etc
x=1 then 21/4 leaves a remainder of 1
x=3 then 63/4 leaves a remainder of 3
x=5 then 105/4 leaves a remainder of 1
x=7 then 147/4 leaves a remainder of 3 ETC
So even combined both statements are INSUFFICIENT. The Answer is E. Definitely not C
(1)yz=3. Putting this value in 7xyz we have
21x where x could be any value eg. 1,2,3,4 etc
When x is 1 21/4 gives a remainder of 1
When x is 2 42/4 gives a remainder of 2
When x is 3 63/4 gives a remainder of 3 etc
No fixed answer INSUFFICIENT
(2)x is odd
7x=ODD for sure as both are odd
yz could be odd or even
If both are odd then 7xyz is odd so remainder could be 1,2,3
If both are even then 7xyz is even(as Even*Even=Even)
And if 7xyz is even then remainder could be 0 or 2 INSUFFICIENT
Combining both we have
21x and that x is odd. So 21x is odd i.e. 7xyz is odd
x could be 1,3,5,7 etc
x=1 then 21/4 leaves a remainder of 1
x=3 then 63/4 leaves a remainder of 3
x=5 then 105/4 leaves a remainder of 1
x=7 then 147/4 leaves a remainder of 3 ETC
So even combined both statements are INSUFFICIENT. The Answer is E. Definitely not C
not sure if this is what question meant....I thought these are different digits
If my post helped you- let me know by pushing the thanks button. Thanks
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joshmachine440
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GMATinsight
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Considering that 7xyz is product of 4 numbers 7 x X x Y x Z [GMAT usually mentions a number like 7XYZ as '4-Digit Number 7XYZ' if it expects us to consider it as 4 digit number]leonswati wrote:If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?
1) yz = 3
2) x is odd.
Plz help me solve it....
Question : what is the remainder when 7 x X x Y x Z is divided by 4?
Statement 1) YZ = 3
X is unknown therefore INSUFFICIENT
Statement 2) x is odd.
Y and Z are unknown therefore INSUFFICIENT
Combining the two statements
7 x X x Y x Z = = 7 x (3) x X = 21X
For X = 1, Remainder when No. is divided by 4 = 1
For X = 3, Remainder when No. is divided by 4 = 3
INSUFFICIENT
Answer: Option E
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GMATinsight
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Considering that 7XYZ is a 4-Digit numbers 7 x X x Y x ZIf x, y, and z are all positive integers, what is the remainder when a 4-Digit Number 7XYZ is divided by 4?
1) Product of Y and Z = 3
2) X is odd.
Question : what is the remainder when 7XYZ is divided by 4?
Divisibility Test of 4 : If last two digits of a Number are divisible by 4 then the number will be divisible by 4
Statement 1) YZ = 3 [Considering that product of Y and Z is 3)
Y = 3 and X = 1, 7XYZ will leave a remainder 3 when divided by 4
Y = 1 and X = 3, 7XYZ will leave a remainder 3 when divided by 4
INSUFFICIENT
Statement 2) X is odd.
Y and Z are unknown therefore INSUFFICIENT
Combining the two statements
That's not beneficial as we only have to know about Y and Z to answer the question
INSUFFICIENT
Answer: Option E
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Brent@GMATPrepNow
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If joshmachine440 is correct (see 3 posts above), then the question SHOULD read as follows:
Let's first examine some powers of 7
7¹ = 7 When we divide 7 by 4, the remainder is 3
7² = 49 When we divide 49 by 4, the remainder is 1
7³ = 343 When we divide 343 by 4, the remainder is 3
7� = 2401 When we divide 2401 by 4, the remainder is 1
etc.
So, when 7^(ODD integer) is divided by 4, the remainder is 3
When 7^(EVEN integer) is divided by 4, the remainder is 1
So, the goal here is to determine whether xyz is guaranteed to be either ODD or EVEN.
Statement 1: yz = 3
This does not guarantee that xyz is odd, nor does it guarantee that xyz is even.
Consider these two CONFLICTING cases:
Case a: x = 1, y = 1 and z = 3. Here, xyz = 3, which is ODD. In the case 7^(xyz) divided by 4 leaves remainder 3
Case b: x = 2, y = 1 and z = 3. Here, xyz = 6, which is EVEN. In the case 7^(xyz) divided by 4 leaves remainder 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x is odd.
This does not guarantee that xyz is odd, nor does it guarantee that xyz is even.
Consider these two CONFLICTING cases:
Case a: x = 1, y = 1 and z = 1. Here, xyz = 1, which is ODD. In the case 7^(xyz) divided by 4 leaves remainder 3
Case b: x = 1, y = 1 and z = 2. Here, xyz = 2, which is EVEN. In the case 7^(xyz) divided by 4 leaves remainder 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that yz = 3 = ODD
Statement 2 tells us that x is ODD
So, xyz = (x)(yz) = (ODD)(ODD) = ODD
So, we can be certain that xyz is ODD, which means 7^(xyz) divided by 4 leaves remainder 3
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
Target question: What is the remainder when 7^(xyz) is divided by 4If x, y, and z are all positive integers, what is the remainder when 7^(xyz) is divided by 4?
1) yz = 3
2) x is odd.
Let's first examine some powers of 7
7¹ = 7 When we divide 7 by 4, the remainder is 3
7² = 49 When we divide 49 by 4, the remainder is 1
7³ = 343 When we divide 343 by 4, the remainder is 3
7� = 2401 When we divide 2401 by 4, the remainder is 1
etc.
So, when 7^(ODD integer) is divided by 4, the remainder is 3
When 7^(EVEN integer) is divided by 4, the remainder is 1
So, the goal here is to determine whether xyz is guaranteed to be either ODD or EVEN.
Statement 1: yz = 3
This does not guarantee that xyz is odd, nor does it guarantee that xyz is even.
Consider these two CONFLICTING cases:
Case a: x = 1, y = 1 and z = 3. Here, xyz = 3, which is ODD. In the case 7^(xyz) divided by 4 leaves remainder 3
Case b: x = 2, y = 1 and z = 3. Here, xyz = 6, which is EVEN. In the case 7^(xyz) divided by 4 leaves remainder 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x is odd.
This does not guarantee that xyz is odd, nor does it guarantee that xyz is even.
Consider these two CONFLICTING cases:
Case a: x = 1, y = 1 and z = 1. Here, xyz = 1, which is ODD. In the case 7^(xyz) divided by 4 leaves remainder 3
Case b: x = 1, y = 1 and z = 2. Here, xyz = 2, which is EVEN. In the case 7^(xyz) divided by 4 leaves remainder 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that yz = 3 = ODD
Statement 2 tells us that x is ODD
So, xyz = (x)(yz) = (ODD)(ODD) = ODD
So, we can be certain that xyz is ODD, which means 7^(xyz) divided by 4 leaves remainder 3
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
