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# If A,BCD is a four-digit positive integer such that A,BCD is equal to

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Joined: 02 Sep 2009
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If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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30 Aug 2015, 10:52
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Difficulty:

95% (hard)

Question Stats:

42% (02:48) correct 58% (02:50) wrong based on 161 sessions

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If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Kudos for a correct solution.

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Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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01 Sep 2015, 23:37
5
2
Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Kudos for a correct solution.

Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example.

101 * 10 = 1010

AD = 10 - a factor of 1010
CD = 10 - a factor of 1010
CA = 11 - not a factor of 1010

Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.

10C0 = 1000 + C0 = 999 + C1
C1 is divisible by C1 but we don't know whether 999 is divisible by C1.
999 = 3^3*37
It has no two digit factor ending in 1 so it will not be divisible by C1.

In case, 1s and 0s don't come to mind, here is how you could arrive at 10C0:

A bit of visualisation will help:

........A B C
........x A B
__________________
........M N P
.....Q R S x
__________________
.....A B C D

Note the leftmost digits: A*A gives you Q and you get A in the final product.
1*1 = 1 (possible value of Q and A)
2*2 = 4 (even if there were a carry over from previous calculation of M + R, Q+1 = 5, not A(2)). Hence A = 2 is not possible.
3*3 = 9 (even if there were a carry over from previous calculation, Q+1 would be 10 and the final product would be a 5 digit number.)

A cannot be any larger value because then the product would be a 5 digit number. Hence A must be 1 and there must be NO carryover from the previous calculation.

........1 B C
........x 1 B
__________________
........B N P
.....1 B C x
__________________
.....1 B C D

Now note that B+B = B. This is only possible if B = 0 or B = 9 (with a carryover from previous addition). But 9 + 9 + 1 = 19 and that would give a carryover to the next addition which we don't want. So B must be 0

So you get 10C * 10 = 10C0

This is true for all values of C from 1 to 9. (C cannot be 0 because CD is a two digit number)
_________________
Karishma
Veritas Prep GMAT Instructor

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Joined: 14 Mar 2014
Posts: 146
GMAT 1: 710 Q50 V34
If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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31 Aug 2015, 03:20
4
1
Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Kudos for a correct solution.

IMO : C

AB=10*A+B
ABC=100*A+10*B+C
ABCD=1000*A+100*B+10*C+D
AB*ABC=1000A^2 +100AB+10AC+100AB+10B^2+BC
=ABCD=1000*A+100*B+10*C+D
SO,A^2=A SO A=1
100AB=100B
10AC=10C
SO D=100AB+10B^2+BC
This is possible only if B=0 SO,D=0
Hence A=1,B=0,D=0 C can be any digit.
From options AD=10 which is always a factor of ABCD since D=0
CD may or may not be a factor ( if C = 5 then yes, if 6 then no)

CA cannot be a factor of ABCD. (If c = 5 then 51 not a factor of 1050)

So only CA cant be the factor.CD may be factor.
_________________
I'm happy, if I make math for you slightly clearer
And yes, I like kudos
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##### General Discussion
Manager
Joined: 05 Jun 2015
Posts: 78
Location: United States
WE: Engineering (Transportation)
Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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01 Sep 2015, 19:46
ABCD = ABC * AB
1000A + 100B + 10C + D = (100A + 10 B + C)* (10A + B)
= 1000 A^2 + 100 (2AB) + 10( B^2 +AC) + BC
=> 1000 (A^2 - A) + 100 (2AB - B) + 10 ( AC + B^2 - B) + (BC - D) = 0
Given, A >0 ans as B, C , D are single digit integers which means(BC- D) will be atmost two digit number, for LHS to be zero, as the first two co-efficients are positive they need to be zero ( A^2 >=A & 2AB > B for all values).
2AB = B => A = 1/2 or B = 0.. As A is integer , B= 0
A^2 - A = 0 => A = 1

As D is product of B & C => D is 0

The number is of the form 10C0
Any value of C would be satisfy

AD (10 ) is factor of 10C0
CD (C0) if C is 3, 30 is not factor of 1030
CA (C1) if C is 1, 11 is not factor of 1010

option D
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Joined: 01 Oct 2016
Posts: 1
Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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27 Jul 2017, 09:04
VeritasPrepKarishma wrote:
Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Kudos for a correct solution.

Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example.

101 * 10 = 1010

AD = 10 - a factor of 1010
CD = 10 - a factor of 1010
CA = 11 - not a factor of 1010

Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.

10C0 = 1000 + C0 = 999 + C1
C1 is divisible by C1 but we don't know whether 999 is divisible by C1.
999 = 3^3*37
It has no two digit factor ending in 1 so it will not be divisible by C1.

In case, 1s and 0s don't come to mind, here is how you could arrive at 10C0:

A bit of visualisation will help:

........A B C
........x A B
__________________
........M N P
.....Q R S x
__________________
.....A B C D

Note the leftmost digits: A*A gives you Q and you get A in the final product.
1*1 = 1 (possible value of Q and A)
2*2 = 4 (even if there were a carry over from previous calculation of M + R, Q+1 = 5, not A(2)). Hence A = 2 is not possible.
3*3 = 9 (even if there were a carry over from previous calculation, Q+1 would be 10 and the final product would be a 5 digit number.)

A cannot be any larger value because then the product would be a 5 digit number. Hence A must be 1 and there must be NO carryover from the previous calculation.

........1 B C
........x 1 B
__________________
........B N P
.....1 B C x
__________________
.....1 B C D

Now note that B+B = B. This is only possible if B = 0 or B = 9 (with a carryover from previous addition). But 9 + 9 + 1 = 19 and that would give a carryover to the next addition which we don't want. So B must be 0

So you get 10C * 10 = 10C0

This is true for all values of C from 1 to 9. (C cannot be 0 because CD is a two digit number)

if we put 107 * 10 = 1070 , in this case C is 7 and D is 0
then 1070 is not divisible by 70
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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27 Jul 2017, 22:11
optimusprime07 wrote:
if we put 107 * 10 = 1070 , in this case C is 7 and D is 0
then 1070 is not divisible by 70

Yes, but there exists a case in which it is. In the case of 101*10 = 1010, C = 1 and D = 0.
1010 is divisible by 10.
But the question asks: "which of the following two-digit numbers cannot be a factor of A,BCD?"
To qualify, CD should not be a factor of ABCD for all values of CD.
_________________
Karishma
Veritas Prep GMAT Instructor

Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to   [#permalink] 27 Jul 2017, 22:11
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