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The digits of a four-digit positive integer have a sum of 19. If the

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Bunuel
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The digits of a four-digit positive integer have a sum of 19. If the [#permalink] New post   10 Jan 2018, 23:32
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The digits of a four-digit positive integer have a sum of 19. If the ten's digit is twice the units digit and the hundreds digit is 8/3 the units digit, what is the number?

A. 863
B. 1842
C. 2863
D. 8263
E. 9820
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Re: The digits of a four-digit positive integer have a sum of 19. If the [#permalink] New post   11 Jan 2018, 00:20
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Four digit number = abcd

a+b+c+d = 19

c=2d

b=8/3d

d can have values 3,6,9

since all are single digit , d can only be 3 or else in case 6 and 9 , b will be 16 and 24

when d =3 , b =8

c=6

and from given sum a will b so number would be 2863
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The digits of a four-digit positive integer have a sum of 19. If the [#permalink] New post   11 Jan 2018, 03:03
Bunuel wrote:
The digits of a four-digit positive integer have a sum of 19. If the ten's digit is twice the units digit and the hundreds digit is 8/3 the units digit, what is the number?

A. 863
B. 1842
C. 2863
D. 8263
E. 9820


We are given the following attributes of the number
1. 4 digit number
2. Sum of the digits is 19
3. ten's digit is twice one's digit
4. 3 times the hundreds digit is 8 times the units digit

Eliminating wrong answer options:
Option A is not a 4 digit number
Option E does not satisfy Pt 3(ten's digit is not twice one's digit)
Option B and D do not satisfy Pt 4(3 times the hundreds digit is 8 times the units digit)

Only Option C remains and 2863 is the correct answer
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Re: The digits of a four-digit positive integer have a sum of 19. If the [#permalink] New post   18 Jan 2018, 08:46
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Bunuel wrote:
The digits of a four-digit positive integer have a sum of 19. If the ten's digit is twice the units digit and the hundreds digit is 8/3 the units digit, what is the number?

A. 863
B. 1842
C. 2863
D. 8263
E. 9820


We can let a, b, c, and d be the units, tens, hundreds, and thousands digits, respectively; thus:

a + b + c + d = 19

and

b = 2a

and

c = 8a/3

We see that the units digit must be a multiple of 3, and since the tens digit is twice the units digit, the only option for the units digit is 3, and thus the tens digit must be 6. Now we can determine c:

c = 8(3)/3 = 8

Finally, we can determine the thousands digit:

3 + 6 + 8 + d = 19

d = 2

Thus, the number is 2,863.

Answer: C
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Re: The digits of a four-digit positive integer have a sum of 19. If the [#permalink]
18 Jan 2018, 08:46
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