math challenge problem need help?

Question

FInd a number so that when the digit 1 is put in front of it, the new number is 5 times the original number. Can you find more than one number?

Answer 1

All the numers of the form 25, 250, 2500, 25000, etc.

Answer 2

To solve it, let's say there is a one-digit number which satistifies such a relation. By putting 1 before that number, we are actually adding ten to it. So

10 + x = 5x, which gives us x = 2.5

Unfortunately, we are thinking of whole numbers here. So we just eliminate the the decimal by multiplying it with 10, and we have 25.

This is all right, because a two-digit number added with a 1 in front is like added to 100. So

100 + x = 5x, with x = 25.

Other answers? All numbers that satisfy the expression 25 x 10^k, where k is any whole number.

Answer 3

Let x be any one of the numbers to be found.

"1" in front of a digit = 10^n,where n is the highest digit position of x.

10^n + x = 5x

Solve for x,
x = 10^n / 4, n can be any positive integer.
n = 1, x = 2.5, 12.5 = 5(2.5)
n = 2, x = 25, 125 = 5(25)
n = 3, x = 250, 1250 = 5(250)
...
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More thoughts:

Actually, n=0 works too if 0.25 can be written as .25.
5(.25) = 1.25

But n<0 doesn't work anyway because
5(.025) = .125 where "1" is not in front of ".".

Answer 4

Suppose the number x is n digits long. This statement is then equivalent to saying that 10^n + x = 5x. This in turn means that 10^n=4x, so this is possible if and only if 10^n is divisible by 4, which in turn is possible if and only if n is at least 2. In that case, we have that our solution will be 10^n/4. So possible solutions are:

25*5 = 125
250*5 = 1250
2500*5 = 12500

And so on.

Answer 5

maybe u should try specifying the range of number(tens?hundreds?,thousands? etc that we are looking at here.
if u were to look at the tens(10-99),then adding 1 in front of that number will make your new number(ie after addin the 1 in front) to be a hundred(range is 100-999).i.e lets say our original number was x where 10x=25.i.e if u add 1 in front of 25,our new number will be 125 which is 5 times 25.u can continue doing so until u find a number in each class i.e if it existsd one infront then new number has to be thousand etc..hope that helps.

Answer 6

It will be 25 because when you stick 1 in front of it you get, 125 which is equal to 25 times 5.

Answer 7

25 > 5*25=125

Answer 8

2.5 and 12.5
25 and 125
250 and 1250
2500 and 12500
25000 and 125000
etc.

I came up with a system of equations.
If x is the smaller number and y is the bigger, then

5x = y

and based on place value

x + 10 = y
x + 100 = y
x + 1000 = y
etc.

and found the intersection of 5x = y and the place value graph on my caluclator, which indicates a pattern that the answer is always a power-of-ten multiple of 25 and 125.

Fun, thanks!

Answer 9

25. 125:25=5

Answer 10

ok we could see... show that for each non-damaging integer n, the extensive sort 7^7^n +a million is the fabricated from a minimum of 2n+3 (not inevitably different) top aspects. This replace into from the 2007 USAMO. on the competition, I in basic terms wrote for n=0, the difficulty is trivial. something of the data by skill of induction is left as an exercising for the grader. LOL

Answer 11

The numbers are 0.0025, 0.25, 2.5, 25,250,2500,25000, 250000, etc.
Basically any number that can be written as 2.5*10^n
where n is an integer.