 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Duplicate Digit Determination IV (Posted on 2011-06-05) (I) Each of x and y is a positive integer with x < y such that, reading from left to right, the first two digits in the base ten expansions of 1978x and 1978y are congruent.

Determine the minimum value of x+y.

(II) What is the minimum value of x+y - if, keeping all the other conditions in (i) unaltered, the first three digits in the base ten expansions of 1978x and 1978y are congruent?

Note: None of the expansions of 1978x and 1978y can contain any leading zero.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
For 1978^x and 1978^y to start with the same digits then 1978^(y-x) will start with 99... or 100...

1978^27 starts with 995...  Then 1978^(x+27) will have similar first digits to 1978^x.

x=1 makes 1978^1 equals 1978 and 1978^28 begin with 1969....  These match for the first two digits, so x+y=29 for part 1.

x=17 makes 1978^17 begin with 1086... and 1978^44 begin with 1081...  These match for the first three digits, so x+y=61 for part 2.

 Posted by Brian Smith on 2017-07-02 14:14:56 Please log in:

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