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5 Digit Number (Posted on 2003-08-20) Difficulty: 3 of 5
I am thinking of a 5 digit number that when tripled is a perfect square.

Also, when the 5 digit number is split, the first number is double the second one. What is the five digit number?

(Splitting the 5 digit number into two numbers means 12345 into 1 and 2345 or 123 and 45.)

See The Solution Submitted by Gamer    
Rating: 3.0000 (6 votes)

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Solution Puzzle Solution | Comment 10 of 13 |
(In reply to Answer by K Sengupta)

Let us denote the 5 digit number by abcde, where each of the digits b, c, d and e denote a decimal digit from 0 to 9 and 'a' denotes a decimal digit from 1 to 9.

Then, the split can be effected in two ways:

(i) abc = 2*de, where c is non zero, and:

(ii) ab = 2*(0de), where c = 0

For the latter case, ab0de = 1000*ab + e = 2001*de

Thus, it follows that if:

abcde = 3*(x^2), whenever x is an integer, then we must have:
2001*de = 3*x^2
or, x^2 = 6667*de = 29*23*de

Accordingly, fot the rhs to be a perfect square, it follows that the minimum value of de is 667, since de = 00 would yield abc=000, giving leading zeroes, which is a contradiction. But, de can contain a maximum of two digits, while 667 is a 3 digit number. This is leads to a contradiction.

For the former case, we have:

abcde = 100*abc + de = 210*de

Then, abcde = 3*(x^2) gives:

201*de = 3(x^2)
or, 67*de = x^2

Since, 67 is a prime, it follows that:

67 must divide x, so that x=67*y, for some integer y.

Thus, 67*de = (67^2)*(y^2)
or, de = 67*(y^2)

In this situation, y=0 gives de = 00, so that abc=00. But, a=0 would give at least one leading zero and thus is a contradiction.

If y>=2, then de>= 248, which is a contradiction.

Thus, y=1, so that de=67, giving abc = 134

Consequently, the required 5 digit number is 13467.


134 = 67*2

13467*3 = (67)*(201)*3 = (67)*(67*3)*(3) = (67*3)*(67*3) = (201)*(201) = (201)^2 

  Posted by K Sengupta on 2008-05-19 04:42:20
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