Becoming Number 1

By knowing that a and b are integers between 0 and 9 inclusive, we can know that the maximum value of $a \times b$ is 81, so we only need to check 1, 11, 21, 31, 41, 51, 61, 71, and 81

By that, we can find $1 \times 1 = 1$ $9 \times 9 = 81$ $3 \times 7 = 21$ $7 \times 3 = 21$

Hence, we know that the answer is 4 ordered pairs

First off, if you multiply an even number (a number that ends in an even digit) by any number, the answer will be even, so that rules out even values for $a$ and $b$ .

Also, the last digit of the product of any odd number and $5$ is also $5$ , which eliminates $5$ from the set of possible values, and we are left with $1$ , $3$ , $7$ and $9$ .

From here it can easily be seen that $1 \times 1 = 1$ , $3 \times 7 = 7 \times 3 = 21$ and $9\times 9 = 81$ , so the ordered pairs $(1,1)$ , $(3,7)$ , $(7,3)$ , and $(9,9)$ satisfy the required property as well.

The solutions are

1x1

3x7

7x3

9x9

So 4

Note that, if either of $a, b$ were even, the product would have an even units digit and could not be $1$ . Similarly, note that if $a = 5$ and/or $b = 5$ , the units digit of $a \times b$ would have to be either $0$ or $5$ . Thus, $a, b \in \{ 1, 3, 7, 9 \}$ . It is easy to see that, for each $a$ , there is a unique $b$ between $0$ and $9$ , inclusive, that satisfies the relation $a \times b \equiv 1 \pmod {10}.$ Specifically, we have that the paris $(1, 1), (3, 7), (7, 3),$ and $(9, 9)$ are the only ones that work. So, our answer is $\boxed {4}$ .

To get 1 as units digit by multiplying any two numbers (between 0 and 9) we could have only these possibilities - $1\times1, 3\times7, 9\times9$

Thus, the only ordered pairs $(a,b)$ are - $(1,1),(3,7),(7,3),(9,9)$

Hence only $\boxed{4}$ ordered pairs.

Obviously, we cannot have either even $a$ or even $b$ .

We cannot choose $5$ too because product of $5$ with any other number gives $5$ or $0$ as the unit digit.

Hence, we are left with only $1,3,7$ and $9$ .

When

$\displaystyle a=1, \, \, b=1$

$\displaystyle a=3, \, \, b=7$

$\displaystyle a=7, \, \, b=3$

$\displaystyle a=9, \, \, b=9$

Therefore, number of ordered pairs are $\fbox{4}$ .

From definition, a and b are single digit numbers. $ab = 10k+1$ => both a and b are odd numbers. Searched in the domain of $1,3,5,7,9$ and found the pairs $(1,1), (3,7), (7,3), (9,9)$

Neither a nor b can be even as then a \times b would become even. None of them can be 0 as then the result would become 0 and none of them can be 5 as multiples of 5 have either 5 or 0 in their units digit. So we check only four odd digits for a viz 1,3,7,9 and get four following ordered pairs of integers: (1, 1), (3, 7), (7, 3), (9, 9)

Let $0<=a<=9$ and $0<=b<=9$

Here we get the units digit 1 when $a \times b$ are $1 \times 1 = 1; 3 \times 7 = 21; 7 \times 3 = 21; 9 \times 9 = 81$

So, there is $\boxed{4}$ ordered pairs of integers of $(a, b)$ .

there are only four pairs (1,1)(9,9),(7,3),(3,7). we can easily omit even nos. as theis multiply can't result in odd number producing 1 on unit place...