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# In the sum above, each letter represents a digit from 0 to 9

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In the sum above, each letter represents a digit from 0 to 9 [#permalink]

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15 Feb 2008, 02:51
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3 v y v
+ v y v z
_____________
c b c d

In the sum above, each letter represents a digit from 0 to 9. What is the value of c if z = v + 3 and y = v - 3?

(1) b < c
(2) z > 7
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15 Feb 2008, 03:29
I don't see 'a' anywhere in the question. I assume the result is 'a b c d'?

--
[3*1000 + 100*v + 10*y + v] + [1000*v + 100*y + 10*v + z]
= 2673+1222v

B) z > 7 implies v can be 5 or 6 (with 6 we have thou digit > 9) - SUFF

A) b < c - results in more than one value for a

--
result will be 8783 if a=c

Last edited by srp on 15 Feb 2008, 10:19, edited 1 time in total.
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15 Feb 2008, 03:41
This is a tough 'un. First of all I'm guessing the sum is abcd, not cbcd as you wrote.

(1) b<c, Since the sums that lead to b and c are identical (y+v in the tens place and v+y in the hundreds place), the only way that this can happen is if the sum in the units place (v+z) is at least 10, and y+v is less than 10. This means that v+z = v+ (v+3) = 2v+3 > 10. Also we know that v+y = v+(v-3)=2v-3 < 10. This doesn't tell us what v is, only that v is between 4 and 6. Not suff.

(2) z>7, since z=v+3 this means that v>4. Again not suff.

Taking them together all we know is that v is either 5 or 6. But there is no way to know exactly what is a=v+3.

I think the answer is (E).
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15 Feb 2008, 04:15
The question is correct as written: there are no a's. Sorry about the earlier version
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15 Feb 2008, 08:09
kevincan wrote:
The question is correct as written: there are no a's. Sorry about the earlier version

sorry again. could you confirm that it's

3vyv
vyvz
----
cbcd

and z = v+3, y = v-3. could you also confirm the question?
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15 Feb 2008, 10:01
3vyv
vyvz
----
cbcd

and z = v+3, y = v-3.

(3*1000 + v*100 + y*10 + v) + (v*1000 + y*100 + v*10 + z) = c*1000 + b*100 + c*10 + d
(3000 + 101v + 10y) + (1010v + 100y + z) = 1010c + 100b + d
(3000 + 101v + 10(v - 3)) + (1010v + 100(v - 3) + v + 3) = 1010c + 100b + d
(2970 + 111v) + (1111v - 297) = 1010c + 100b + d
2673 + 1222v = 1010c + 100b + d

- v cannot be 0, 1, 2 cause y will be negative, v cannot be 7, 8, 9 cause z will be two digits

b < c, information is not needed and not helpful since v can only be digits from 3 to 6

v = 3
3 v y v => 3 3 0 3
v y v z => 3 0 3 6
-------------------
= 6 3 3 9, 6 =/= 3, does not suit cbcd.

v = 4
3 v y v => 3 4 1 4
v y v z => 4 1 4 7
-------------------
= 7 5 6 1, 7 =/= 6, does not suit cbcd.

z > 7, z = 8 or 9 since it's a digit problem
If z = 8, v = 5, y = 2

3 v y v => 3 5 2 5
v y v z => 5 2 5 8
-------------------
= 8 7 8 3, c = 8, suits cbcd, yet c = z.

If z = 9, v = 6, y = 3

3 v y v => 3 6 3 6
v y v z => 6 3 6 9
-------------------
= 1 0 0 0 5, does not work since there's one more digit involved.

What choices do you make if the problem itself is solve-able without either statements -.-?
D or E?
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15 Feb 2008, 10:43
From the information given we have:
d=2v+3
c= 2v-3
b= 2v-3
c=3+v = z
1) b<c => 2v-3<3+v => v<6
We know that c#b, unless z+v>10. From this b+1=c. From this v>3. So the only possible values of v are 4 and 5. For v=4 is impossible, so v=5 - suff
2) z>7 => c>7 => 3+v>7 => v>4. So v could have possible values 5,6,7,8,9. For every v>=6 the result will have 5 digits. So v=5.
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