Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"

Latest revision as of 16:25, 11 July 2021

Problem

What is the leftmost digit of the product $\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }}?$

Solution

We notice that $16000\cdots \times 25000\cdots = 16 \times 25 \times 10^{198} = 400 * 10^{198}$ In addition, we notice that $16200\cdots \times 25300\cdots = 162 \times 253 \times 10^{194} = 40986 \times 10^{194}$

Since $16000\cdots \times 25000\cdots < \underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }} < 16200\cdots \times 25300\cdots$

We conclude that the leftmost digit must be $\boxed{4}$ .

~Bradygho

Solution 2

By multiplying out $16 \cdot 25$ , $161 \cdot 252$ , and $1616 \cdot 2525$ , we notice that the first $2$ digits don't change even when we continue to add more digits. With this observation, we can conclude that the first digit of the product is $\boxed{4}$ .

~Mathdreams

Solution 3

Remove factors of $16$ and $25$ to get $\left(\underbrace{101010101 \cdots}_{\text{50 0s and 50 1s}} \right)^2 \cdot 400$ . Recall by Pascal's triangle that $11=121$ , $101=10201$ , so the leftmost digit is guaranteed to be $1$ . Now, multiplying by our scale factor the answer is $\boxed{4}$ $\linebreak$ ~Geometry285

@@ Line 25: / Line 25: @@
 <math>\linebreak</math>
 ~Geometry285
 ==See also==
-#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
+#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
-#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
+#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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