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 Chop it off (Posted on 2007-03-27)
Which integers are such that if the units digit is chopped off, the original number is a multiple of the new one?

 No Solution Yet Submitted by e.g. Rating: 3.0000 (1 votes)

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 Solution To The Problem Comment 5 of 5 |

It can be readily seen that when the last digit of a number is
deleted, the number decreased by not less than 10 times. A number which decreases exactly 10 times when its last digit is deleted must have a zero at the end. Accordingly, all such numbers of the form 10*p, where p is a positive integer will satisfy conditions of the problem.

Now, to investigate the numbers wose units digits are Non Zero. Let us suppose that a whole number decreases more than 10 times when its last digit is deleted, so that; we can let it decrease (10+a) times (whenever d> =1). Let y be the quotient resulting from the division of the number x by 10 and let z be the digit in the units place of the number x. Then, x = 10y + z. After the last digit of the number x is deleted, we obtain the number y. Accordingly, by conditions of the problem, we obtain x = (10+a)y; or, : 10y+z = (10+a)y, giving z = ay. Since z< 10, we must have y< 10 and: a< 10. Accordingly, the numbers possessing the required property( and NOT having a ZERO at the end)  possess only two digits and can only decrease not more than 19 times when its last digit is deleted.

When 10+a = 11, then a=1,
giving z = y and x = 11y, where y=1,2,...., 9.
Hence, the numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99.

When 10+a = 12, then a=2,
giving z = 2y and x = 12y, where 2y< 10, so that y=1,2,3,4.
Hence, the numbers are 12, 24, 36, 48

When 10+a = 13, then a=3,
giving z = 3y and x = 13y, where 3y< 10, so that y=1,2,3
Hence, the numbers are 13, 26, 39

When 10+a = 14, then a=4,
giving z = 4y and x = 14y, where 4y< 10, so that y=1,2
Hence, the numbers are 14, 28

When 10+a = 15, then a=5,
giving z = 5y and x = 15y, where 5y< 10, so that y=1
Hence, the number is 15

When 10+a = 16, 17, 18, 19 then a=6, 7, 8, 9, we have y=1, so thatthe numbers are 16, 17, 18, 19

Thus, the integers satisfying conditions of the problem :

EITHER possesses  the form 10*p for any positive integer p.

OR, is equal to any of the following:

11,12,13,14,15,16,17,18,19, 22,24,26,28, 33, 36, 39, 44,
48, 55, 66, 77, 88, 99

Edited on April 11, 2007, 4:04 pm
 Posted by K Sengupta on 2007-04-11 05:52:01

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