Maths question regarding factors

[ujjwal3097]

hi everyone
this was the question that my teacher gave me yesterday and answered it but she didn't told me whether i was correct or not so please help me out .
Question:Digit number with factors as 1 and the number itself and any other number.::::::::
Answer: 25

 

[axmls]

I'm not exactly sure what the question is here. Do you mind rephrasing it? The number 25 does not have only 1 and itself as factors.

 

[ujjwal3097]

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I'm not exactly sure what the question is here. Do you mind rephrasing it? The number 25 does not have only 1 and itself as factors.

question says that mention a two digit positive number that has 1 and number itself as a factor plus any other number as a factor that means total 3 factors are there any other number other than 25 .

 

[axmls]

Well, if you mean three distinct factors (1, itself, and another number), then 49 also falls under that category.

 

[pwsnafu]

To show that the only two digit numbers with three distinct factors is 25 and 49, let $x$ be a number with this property. Then clearly $1 | x$ and $x|x$. Let the third factor be $a$. Define $b = x / a$. But if there are only three distinct factors we have $a = b$ so $x = a^2$. Now because $c | a \implies c|x$, so we must have $a$ being a prime number. Then only prime squares between $10 \leq x \leq 99$ are $5^2 = 25$ and $7^2 = 49$.

 

[ujjwal3097]

To show that the only two digit numbers with three distinct factors is 25 and 49, let $x$ be a number with this property. Then clearly $1 | x$ and $x|x$. Let the third factor be $a$. Define $b = x / a$. But if there are only three distinct factors we have $a = b$ so $x = a^2$. Now because $c | a \implies c|x$, so we must have $a$ being a prime number. Then only prime squares between $10 \leq x \leq 99$ are $5^2 = 25$ and $7^2 = 49$.

sir can you please explain the answer in simpler terms I will really appreciate this
thank you

 

[jtbell]

What is the first point that you don't understand?

 

[pwsnafu]

First, we know that 1 and x are factors of x by definition. If x has three distinct factors, then call the last one $a$. We need to show two things: a is prime, and a² is equal to x.

To show that it is prime, suppose that it was composite. So there is another number, $c$, which is a factor of $a$. But $c$ itself is a factor of x. Hence c must divide x, which contradicts x having only three factors. Hence a is a prime number.

To show the latter, we know $a$ divides $x$. Define b = a divided by x. So ab = x. But this means b is also a factor of x. If x has three factors, the only candidates of b are 1, a, and x. b=1 means a = x, which is a contradiction. Similarly b can't be x. So the only remaining possibility is a = b, and so $x=a^2$

 

[HallsofIvy]

Ah- I missed the requirement that the number be a two digit number. If a number has two distinct prime factors, say a, b, then it has 1, a, b, and ab as factors so to have only 1, itself, and one other number as factors, It must be of the form $a^2$ where a is a prime number- that has only 1, a and $a^2$ as factors. To be two digits, a must be larger than 3 (because $3^2= 9$ which has only one digit) and less than 11 (because $11^2= 121$ which has three digits).