1999 CEMC Pascal Problems/Problem 20

Problem

The numbers $49, 29, 9, 40, 22, 15, 53, 33, 13, 47$ are grouped in pairs so that the sum of each pair is the same. Which number is paired with $15$ ?

$\text{ (A) }\ 6 \qquad\text{ (B) }\ 5 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 1 \qquad\text{ (E) }\ 0$

Solution 1

We can notice that the units digit of $5$ raised to any power is $5$ since $5 \times 5 = 25$ , which has a units digit of $5$ .

Thus, the units digit of the expression given is the same as the units digit of:

$(5 + 1)(5 + 1)(5 + 1)(5 + 1) = 6^4$

We can also notice that the units digit of $6$ to any power is $\boxed {\textbf {(A) } 6}$ using similar logic with $6 \times 6$ having a units digit of $6$ .

~anabel.disher

Solution 1.1

We can use the same logic to find that the units digit of the expression given is the same as the units digit of $6^4$ .

However, we can solve $6^4$ to get $1296$ , which has a units digit of $\boxed {\textbf {(A) } 6}$ .

~anabel.disher

Solution 2 (unrecommended)

We can solve for the value of the expression, and then use its ones digit: $(5 + 1)(5^3 + 1)(5^6 + 1)(5^{12} + 1)$

$=6 \times (125 + 1) \times (15625 + 1) \times (244140625 + 1)$

$=6 \times 126 \times 15626 \times 244140626$

$=2,884,095,714,938,256$

We then can see that the units digit is $\boxed {\textbf {(A) } 6}$ .

However, this method is not recommended, and takes a lot of computation.

~anabel.disher

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1999 CEMC Pascal Problems/Problem 20

Contents

Problem

Solution 1

Solution 1.1

Solution 2 (unrecommended)