Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Question

Homi · Accepted Answer

We will use the concept of remainders to write algebraic expressions for the number and then find a common pattern.

Step 1 — Expressing the conditions Let the number we are looking for be N. When N is divided by 3, the remainder is 2. This means N can be written as 3 multiplied by some whole number, plus 2. Let this whole number be $k$ .

$N = 3k + 2$

When N is divided by 4, the remainder is 2. This means N can also be written as 4 multiplied by some other whole number, plus 2. Let this other whole number be $m$ .

$N = 4m + 2$

Step 2 — Finding the first few numbers Let us list numbers that fit the first condition. If $k=0$ , $N = 3(0) + 2 = 2$ . If $k=1$ , $N = 3(1) + 2 = 5$ . If $k=2$ , $N = 3(2) + 2 = 8$ . If $k=3$ , $N = 3(3) + 2 = 11$ . If $k=4$ , $N = 3(4) + 2 = 14$ . If $k=5$ , $N = 3(5) + 2 = 17$ . If $k=6$ , $N = 3(6) + 2 = 20$ . If $k=7$ , $N = 3(7) + 2 = 23$ . If $k=8$ , $N = 3(8) + 2 = 26$ . So, numbers like this are 2, 5, 8, 11, 14, 17, 20, 23, 26, ...

Now, let us list numbers that fit the second condition. If $m=0$ , $N = 4(0) + 2 = 2$ . If $m=1$ , $N = 4(1) + 2 = 6$ . If $m=2$ , $N = 4(2) + 2 = 10$ . If $m=3$ , $N = 4(3) + 2 = 14$ . If $m=4$ , $N = 4(4) + 2 = 18$ . If $m=5$ , $N = 4(5) + 2 = 22$ . If $m=6$ , $N = 4(6) + 2 = 26$ . So, numbers like this are 2, 6, 10, 14, 18, 22, 26, ...

We look for numbers that appear in both lists. The common numbers are 2, 14, 26. Let us find a few more by continuing the pattern. The difference between these numbers is $14 - 2 = 12$ and $26 - 14 = 12$ . So, the next numbers will also have a difference of 12. $26 + 12 = 38$ . $38 + 12 = 50$ .

$\boxed{\text{Some numbers are 2, 14, 26, 38, 50}}$

Diagram 1

Step 3 — Finding the general algebraic expression The number N must satisfy both conditions. So, the two expressions for N must be equal.

$3k + 2 = 4m + 2$

We can subtract 2 from both sides of the equation.

$3k = 4m$

This equation tells us that $3k$ must be a multiple of 4. Since 3 and 4 do not share any common factors other than 1 (they are coprime), $k$ itself must be a multiple of 4. Let $k$ be equal to 4 multiplied by some other whole number, say $t$ .

$k = 4t$

Now, we substitute this value of $k$ back into our first expression for N.

$N = 3k + 2$

$N = 3(4t) + 2$

$N = 12t + 2$

This expression describes all such numbers. Here, $t$ can be any non-negative whole number (0, 1, 2, 3, ...). To match the variable used in the question's final answer, we can replace $t$ with $k$ .

$\boxed{\text{The algebraic expression is } 12k + 2}$

Answer

(i) Some numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4 are 2, 14, 26, 38, 50. (ii) The algebraic expression describing all such numbers is 12k + 2 (where k is a non-negative integer).

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