3087 - a special number. how?

Algebra Level 3

$a, b , c , d \in \mathbb{N}$ , such that ,

$\overline{abcd} - \overline{dcba} = 3087$

Let $a > b > c > d$

If $a = 8$ , find $a + b + c + d$ .

Details and Assumptions -

If $\overline{abcd}$ is $5964$ , then $a = 5 , b = 9 , c = 6 , d = 4 .$

and dcba is 4695 .

The answer is 26.

5 solutions

Guru Prasaadh
Mar 17, 2015

for gods sake dot try wating ur tym seeing the other solutions and by finding values. instead read the question 2 times ur first observation would be that d is 5 so with some iq u should figure out that the rest two numbers should be 6 and 7 as all are integers and a>b>c>d.

I agree. That's quite a very ignored given out there.

Yves Tedera - 6 years, 2 months ago

Danny Nguyen
Jan 27, 2015

Nihar Mahajan
Jan 22, 2015

Let a = d + 3 , b = d + 2 , c = d + 1

abcd = 1000(d+3) + 100(d+2) + 10(d+1) +d .................(1)

dcba = 1000(d) + 100(d+1) + 10(d+2) + d+3 .................. (2)

(1) - (2) gives us ,

abcd - dcba = 1000(3) + 100(1) +10(-1) -3

abcd - dcba = 3000 + 100 -10 -3

abcd - dcba = 3087 .

here , a = 8 .

so d = 5 , b = 7 , c = 6

so , a + b + c + d = 8 + 7 + 6 + 5 = 26.

Isn't it interesting about 3087?

That looks assuming that abcd are in a decreasing AP with common d = 1 or else you don't get the 3087 result

Lucky Kaul - 6 years, 3 months ago

Ashish Gupta
Apr 28, 2015

Abcd - dcba = 3087

Writing abcd = 1000a + 100b + 10c + d and similarly for dcba, We get 111d + 10(c-b) = 545 Now, d can only be 5, and thus b and c are 7 and 6 respectively. Sum comes out to be 26.

Note: By simple observation, d can only be 5 (15 - 8 = 7 = last digit of 3087)

Rush Sgt
Mar 8, 2015

If a=8 and a>b>c>d. Then 8bcd-dcb8=3087. d should be 5 because the last digit on the right side is 7 so 8bc5-5cb8=3087. Knowing all these is obvious that 8765-5678=3087 so a+b+c+d=8+7+6+5=26. The answer then is 26.

3087 - a special number. how?

The answer is 26.

5 solutions

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