Combinations and variations

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Loki123

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How many different four-numbered numbers can be made from 1, 2, 3, 4, 5 so that they can be divided by 4? Numbers can be repeated.

I was given this question in hand-written form so I do not have the exact question or its source unfortunately.

n=5
I think there are 5 cases of numbers in this case that can be divided by 4. I wrote them. Calculated them using variations since placement of the numbers matters.
I got 125, but I was informed the correct answer is 120.
How?IMG_20220411_133400.jpg
 
Loki123 said:
How many different four-numbered numbers can be made from 1, 2, 3, 4, 5 so that they can be divided by 4? Numbers can be repeated.

I was given this question in hand-written form so I do not have the exact question or its source unfortunately.

n=5
I think there are 5 cases of numbers in this case that can be divided by 4. I wrote them. Calculated them using variations since placement of the numbers matters.
I got 125, but I was informed the correct answer is 120.
How?View attachment 32084
Click to expand...
A number

WXYZ

Is divisible by 4, if YZ is divisible by 4.
 
Loki123 said:
How many different four-numbered numbers can be made from 1, 2, 3, 4, 5 so that they can be divided by 4? Numbers can be repeated.

I was given this question in hand-written form so I do not have the exact question or its source unfortunately.

n=5
I think there are 5 cases of numbers in this case that can be divided by 4. I wrote them. Calculated them using variations since placement of the numbers matters.
I got 125, but I was informed the correct answer is 120.
How?View attachment 32084
Click to expand...
I think 125 is correct.
 
Consider a number in the form

[math]p = x_1 * 10^3 + x_2 * 10^2 + x_3 * 10 + x_4 \text { where } 1 \le x_i \le 5 \text { and } x_i \in \mathbb Z.[/math]
Obviously, there are 625 possible values of p. How many are evenly divisible by 4?

[math]\text {Let } q = 10x_1 + x_2 \text { and } r = 10x_3 + x_4.\\ \therefore p = 100q + r \implies \dfrac{p}{4} = 25q + \dfrac{r}{4} \implies \\ 4 \ | \ p \iff 4 \ | \ r.[/math]
There are 25 possible values for r. They fall into 5 ranges. 11-15, 21-25, 31-35, 41-45, 51-55. In each range, 3 values are odd and cannot be divisible by 4. The two even values differ by 2 so exactly 1 of those even values is divisible by 4, namely 12, 24, 32, 44, and 52.

I get 5 * 25 = 125. If 120 is correct, some other condition must be present. Of course, 120 may be a typo.
 
JeffM said:
Consider a number in the form

[math]p = x_1 * 10^3 + x_2 * 10^2 + x_3 * 10 + x_4 \text { where } 1 \le x_i \le 5 \text { and } x_i \in \mathbb Z.[/math]
Obviously, there are 625 possible values of p. How many are evenly divisible by 4?

[math]\text {Let } q = 10x_1 + x_2 \text { and } r = 10x_3 + x_4.\\ \therefore p = 100q + r \implies \dfrac{p}{4} = 25q + \dfrac{r}{4} \implies \\ 4 \ | \ p \iff 4 \ | \ r.[/math]
There are 25 possible values for r. They fall into 5 ranges. 11-15, 21-25, 31-35, 41-45, 51-55. In each range, 3 values are odd and cannot be divisible by 4. The two even values differ by 2 so exactly 1 of those even values is divisible by 4, namely 12, 24, 32, 44, and 52.

I get 5 * 25 = 125. If 120 is correct, some other condition must be present. Of course, 120 may be a typo.
Click to expand...
Okay! Thanks.
 
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