Let (abcd) be a four digit number (each letter represents one digit). If (abcd)+(abc)+(ab)+(a) = 2018, what is the product of a*b*c*d?

1 Answer
George C.
Jun 20, 2018

#64#

Explanation:

Given:

#"abcd"+"abc"+"ab"+"a" = 2018#

Note that:

#"abcd"+"abc"+"ab"+"a" ~~ "abcd"+"abc.d"+"ab.cd"+"a.bcd" = 1.111 * "abcd" #

So:

#"abcd" ~~ 2018/1.111 ~~ 1816.38#

We find:

#1816+181+18+1 = 2016#

So:

#"abcd" = 1816+2 = 1818#

Then:

#"a" * "b" * "c" * "d" = 1 * 8 * 1 * 8 = 64#

Alternative Method

Note that:

#"abcd" + "abc" + "ab" + "a" = "aaaa" + "bbb" + "cc" + "d"#

We find:

#2018-1111 = 907#

#907-888=19#

#19-11=8#

So:

#2018 = 1111+888+11+8 = 1818+181+18+1#

So:

#"abcd" = 1818#

and:

#"a" * "b" * "c" * "d" = 1 * 8 * 1 * 8 = 64#

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