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Intern  B
Joined: 08 Dec 2016
Posts: 32
Location: Italy
Schools: IESE '21
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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1
Data sufficiency always is closer to a RC than to a math problem.
Fact: x,y,z are 3 digit number, so the sum of the hundreds cannot exceed the the 1 digit number
Fact: X= Y+Z

Therefore the question is asking if there is a carry over 1.

Understood this the answer is easy to pick.
Current Student B
Joined: 09 Oct 2016
Posts: 87
Location: United States
GMAT 1: 740 Q49 V42 GPA: 3.49

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2
A.

Took me 30 seconds.

The only way this is possible is if both tens digits are less than 5 or both add up to 9.

Think about it logically, when you are adding, the only way the left digit will change is if the right digits add up to 10 or more.

Ex) 149 + 349 = 498

B) this may or may not be true:
153 + 345 = 498
153 + 351 = 504

Posted from my mobile device
Intern  S
Joined: 11 Sep 2017
Posts: 36
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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2

882 = 541 + 341 tens digit satisfies but ans is not.. don't know where I am wrong
Math Expert V
Joined: 02 Sep 2009
Posts: 59587
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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2
Cheryn wrote:

882 = 541 + 341 tens digit satisfies but ans is not.. don't know where I am wrong

You are getting the same YES answer. The hundreds digit of x, which is 8 in your example, equals to the sum of the hundreds digits of y and z, which are 5 and 3: 5 + 3 = 8.
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Senior Manager  P
Joined: 29 Jun 2017
Posts: 417
GPA: 4
WE: Engineering (Transportation)
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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2
1

x = abc = 100 a + 10b + c
y = def = 100d + 10e + f
z = ghi = 100g + 10h + i

since x=y+z
100a + 10b + c = 100(d+g) + 10(e+h)+ f+i ---------equation alpha
lets consider any 2 numbers of let say 2 digits => 12 and 24 => sum is 36 means unit digit 6 is sum of 2 and 4
therefore
if x=y+z unit digit of x is sum of y's and z's units digit
c= f+i
from 1 10b = 10e+10h
add 10b +c = 10(e+h) + f+i
substitue in equation alpha

100a = 100( d+g)
a=d+g => SUFFICIENT.
Intern  B
Joined: 25 Dec 2017
Posts: 7
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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Bunuel, your explanations are top notch! Thank you so much! Manager  B
Joined: 22 Jan 2018
Posts: 52
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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Bunuel wrote:
The question basically asks whether there is a carry over 1 from the tens place to the hundreds place.

How did you conclude that this is what the question is asking?
Manager  B
Joined: 04 Dec 2015
Posts: 145
WE: Operations (Commercial Banking)
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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I am confused..

My analysis:-

Let us solve using variables..

Let x=abc
Let y=pqr
Let z=efg

we need to show if a=p+e

Since x, y and z are three digit positive integers, we can express them as

X=100a + 10b + c

Y=100p + 10q + r

Z=100e + 10f + g

Since x=y+z (given)

We can write

100a+10b+c = 100p+10q+r + 100e+10f+g

—> 100(a-e-p) + 10(b-q-f) + c-r-g = 0 —(1)

statement 1, b=q+f (given)

Above equation (1) can be written as

100(a-e-p) + 10(q+f - q -f ) = r+g-c

—> 100(a-e-p) = r+g-c ....hence statement 1 is not sufficient as we don’t know if c=r+g

Statement 2 -

C=r+g (given)

Putting the value of c in (1),

100(a-e-p) + 10(b-q-f) + c-r-g=0

—> 100(a-e-p) + 10(b-q-f) = 0

Statement 2 is not sufficient as we don’t really know if b=q+f..

Now statement 1 + statement 2

We have b=q+f and c=r+g

Putting the values of both b and c in (1)

100(a-e-p) = 0

—> a-e-p=0
—> a=e+p

Hence sufficient ie. C is the answer..

Experts! What am I missing? Bunuel chetan2u

Kindly help me!!

Posted from my mobile device

Originally posted by INSEADIESE on 02 Jun 2019, 13:51.
Last edited by INSEADIESE on 02 Jun 2019, 14:02, edited 1 time in total.
Manager  B
Joined: 04 Dec 2015
Posts: 145
WE: Operations (Commercial Banking)
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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I am confused..

My analysis:-

Let us solve using variables..

Let x=abc
Let y=pqr
Let z=efg

we need to show if a=p+e

Since x, y and z are three digit positive integers, we can express them as

X=100a + 10b + c

Y=100p + 10q + r

Z=100e + 10f + g

Since x=y+z (given)

We can write

100a+10b+c = 100p+10q+r + 100e+10f+g

—> 100(a-e-p) + 10(b-q-f) + c-r-g = 0 —(1)

statement 1, b=q+f (given)

Above equation (1) can be written as

100(a-e-p) + 10(q+f - q -f ) = r+g-c

—> 100(a-e-p) = r+g-c ....hence statement 1 is not sufficient as we don’t know if c=r+g

Statement 2 -

C=r+g (given)

Putting the value of c in (1),

100(a-e-p) + 10(b-q-f) + c-r-g=0

—> 100(a-e-p) + 10(b-q-f) = 0

Statement 2 is not sufficient as we don’t really know if b=q+f..

Now statement 1 + statement 2

We have b=q+f and c=r+g

Putting the values of both b and c in (1)

100(a-e-p) = 0

—> a-e-p=0
—> a=e+p

Hence sufficient ie. C is the answer..

Kindly help me!!
Math Expert V
Joined: 02 Aug 2009
Posts: 8281
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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1
I am confused..

My analysis:-

Let us solve using variables..

Let x=abc
Let y=pqr
Let z=efg

we need to show if a=p+e

Since x, y and z are three digit positive integers, we can express them as

X=100a + 10b + c

Y=100p + 10q + r

Z=100e + 10f + g

Since x=y+z (given)

We can write

100a+10b+c = 100p+10q+r + 100e+10f+g

—> 100(a-e-p) + 10(b-q-f) + c-r-g = 0 —(1)

statement 1, b=q+f (given)

Above equation (1) can be written as

100(a-e-p) + 10(q+f - q -f ) = r+g-c

—> 100(a-e-p) = r+g-c ....hence statement 1 is not sufficient as we don’t know if c=r+g

Statement 2 -

C=r+g (given)

Putting the value of c in (1),

100(a-e-p) + 10(b-q-f) + c-r-g=0

—> 100(a-e-p) + 10(b-q-f) = 0

Statement 2 is not sufficient as we don’t really know if b=q+f..

Now statement 1 + statement 2

We have b=q+f and c=r+g

Putting the values of both b and c in (1)

100(a-e-p) = 0

—> a-e-p=0
—> a=e+p

Hence sufficient ie. C is the answer..

Experts! What am I missing? Bunuel chetan2u

Kindly help me!!

Posted from my mobile device

Hi

The reason is that the statement II can be inferred from statement I too.
PQR
EFG
_____
ABC
_____

Now Q+F=B means R+G=C and P+E=A. Let us see why.

Let me cancel out Q,F and B as they are not making any difference on hundreds.
P0R
E0G
____
A0C
____

So nothing comes from tens to hundred, therefore P+E=A

The only way P+E is not equal to A is when something comes from tens Q+F, but we are told that the sum is exactly same as B.
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Manager  B
Joined: 04 Dec 2015
Posts: 145
WE: Operations (Commercial Banking)
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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chetan2u wrote:
I am confused..

My analysis:-

Let us solve using variables..

Let x=abc
Let y=pqr
Let z=efg

we need to show if a=p+e

Since x, y and z are three digit positive integers, we can express them as

X=100a + 10b + c

Y=100p + 10q + r

Z=100e + 10f + g

Since x=y+z (given)

We can write

100a+10b+c = 100p+10q+r + 100e+10f+g

—> 100(a-e-p) + 10(b-q-f) + c-r-g = 0 —(1)

statement 1, b=q+f (given)

Above equation (1) can be written as

100(a-e-p) + 10(q+f - q -f ) = r+g-c

—> 100(a-e-p) = r+g-c ....hence statement 1 is not sufficient as we don’t know if c=r+g

Statement 2 -

C=r+g (given)

Putting the value of c in (1),

100(a-e-p) + 10(b-q-f) + c-r-g=0

—> 100(a-e-p) + 10(b-q-f) = 0

Statement 2 is not sufficient as we don’t really know if b=q+f..

Now statement 1 + statement 2

We have b=q+f and c=r+g

Putting the values of both b and c in (1)

100(a-e-p) = 0

—> a-e-p=0
—> a=e+p

Hence sufficient ie. C is the answer..

Experts! What am I missing? Bunuel chetan2u

Kindly help me!!

Posted from my mobile device

Hi

The reason is that the statement II can be inferred from statement I too.
PQR
EFG
_____
ABC
_____

Now Q+F=B means R+G=C and P+E=A. Let us see why.

Let me cancel out Q,F and B as they are not making any difference on hundreds.
P0R
E0G
____
A0C
____

So nothing comes from tens to hundred, therefore P+E=A

The only way P+E is not equal to A is when something comes from tens Q+F, but we are told that the sum is exactly same as B.

what you mean to say is that cases such as 106 + 204 or 503 + 307 are not possible ie. since b=q+f(statement 1)---> 0<=r+g<10??

am i right ?
Intern  B
Joined: 07 May 2019
Posts: 9
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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GMATPrepNow wrote:
conty911 wrote:
If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

(1) The tens digit of x is equal to the sum of the tens digits of y and z.
(2) The units digit of x is equal to the sum of the units digits of y and z.

Target question: Is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

Notice that there are essentially 3 ways for the hundreds digit of x to be different from the sum of the hundreds digits of y and z
Scenario #1: the hundreds digits of y and z add to more than 9. For example, 600 + 900 = 1500. HOWEVER, we can rule out this scenario because we're told that x, y, and z are three-digit integers
Scenario #2: the tens digits of y and z add to more than 9. For example, 141 + 172 = 313.
Scenario #3: the tens digits of y and z add to 9, AND the units digits of y and z add to more than 9. For example, 149 + 159 = 308

Statement 1: The tens digit of x is equal to the sum of the tens digits of y and z.
This rules out scenarios 2 and 3 (plus we already ruled out scenario 1).
So, it must be the case that the hundreds digit of x equals to the sum of the hundreds digits of y and z
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The units digit of x is equal to the sum of the units digits of y and z.
This rules out scenario 3, but not scenario 2. Consider these two conflicting cases:
Case a: y = 100, z = 100 and x = 200, in which case the hundreds digit of x equals the sum of the hundreds digits of y and z
Case b: y = 160, z = 160 and x = 320, in which case the hundreds digit of x does not equal the sum of the hundreds digits of y and z
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent

Hello Brent,

Couldn't it be perceived that scenario 3 is applicable to statement (1)?

Eric
Intern  B
Joined: 30 Jun 2019
Posts: 7
Location: United States
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

Assume
X= a b c
Y= d e f
Z= g h i
Given: X=Y+Z
Question: is a=d+g?
if e+h>10 so its carry one and a=d+g+1
therefore if e+h<10 so yes, a=d+g; and that's exactly case1 - so sufficient.
Case 2 is not relevant so insufficient.
Manager  B
Joined: 03 Aug 2017
Posts: 92
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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Bunuel wrote:
fozzzy wrote:
If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

(1) The tens digit of x is equal to the sum of the tens digits of y and z.
(2) The units digit of x is equal to the sum of the units digits of y and z.

Is there an alternative approach for this problem?

The question basically asks whether there is a carry over 1 from the tens place to the hundreds place.

Consider the following examples:
(i)
123
234
357

Hello

Here when we add the tens digits 2 and 3 there is no carry over 1 from the tens place to the hundreds place, thus the hundreds digit of x (3) is equal to the sum of the hundreds digits of y (1) and z (2).

(ii)
153
147
300

Here when we add the tens digits 5 and 4 and carry over 1 from the units place, we get 10, so we have carry over 1 from the tens place to the hundreds place, thus the hundreds digit of x (3) does NOT equal to the sum of the hundreds digits of y (1) and z (1).

The first statement implies that there is no carry over 1 from the tens place to the hundreds place, thus the hundreds digit of x is equal to the sum of the hundreds digits of y and z. Sufficient.

The second statement does not provide us with sufficient information about carry over 1 from the tens place to the hundreds place.

Hope it's clear.

Hello Bunuel,
Can u please explain how we can establish from Statement 1 that even the units digit of Y& Z = the units digit of X ? According to me we dont know this , do we ? What if the sum of tens digit of Y & Z = that pf X but units digit of Y& Z add up more than 9 then we cant answer the question with certanity . HEnce i thought Option A on its own is not enough .
Manager  G
Joined: 08 Jan 2018
Posts: 129
If x, y, and z are three-digit positive integers and if x =  [#permalink]

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Let, the digits be:

x = abc
y = pqr
z = xyz

Given: x = y + z

We need to know if a = p + x

(1) b = q + y
That means the sum of q + y will always have to be <10. If it is higher than 10, it will not satisfy this condition.
Ex- x:357, y:123, z:234 ; x:449, y:215, z:234 ;
In all cases a = p + x

Satisfies

(2) c = r + z
That means the sum of r + z will always have to be <10. If it is higher than 10, it will not satisfy this condition.

But this also means we are free to assume p + y > 10, which will, in turn, affect the hundreds digits.

Ex- y:193, z:194, x:387 -> a != p + x
y:123, z:234, x:357 -> a = p + x

It does not satisfy.

VP  V
Joined: 23 Feb 2015
Posts: 1338
Re: If x, y, and z are three-digit positive integers and if x =  [#permalink]

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mbaiseasy wrote:
x = ABC
y = DEF
z = GHI

DEF
+GHI
_____
ABC

Question: Is D + G = A? This is true if there is no carry-over from the tens digits' sum.

1. E + H = B, This means there is no carry over to hundreds position. SUFFICIENT.
2. C + F = I, This means there is no carry over to tens position BUT we do not know if there will be a carry over during the sum of tens. INSUFFICIENT.

Shouldn't be the highlighted part C=F+I or C-F=I?
Thanks__
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