Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
− | By multiplying out <math>16 \cdot 25</math>, <math>161 \cdot 252</math>, and <math>1616 \cdot 2525</math>, we notice that the first <math>2</math> 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 <math>4</math>. | + | By multiplying out <math>16 \cdot 25</math>, <math>161 \cdot 252</math>, and <math>1616 \cdot 2525</math>, we notice that the first <math>2</math> 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 <math>\boxed{4}</math>. |
~Mathdreams | ~Mathdreams |
Revision as of 11:34, 11 July 2021
Contents
[hide]Problem
What is the leftmost digit of the product
Solution
We notice that
In addition, we notice that
Since
We conclude that the leftmost digit must be .
~Bradygho
Solution 2
By multiplying out ,
, and
, we notice that the first
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
.
~Mathdreams
Solution 3
Remove factors of and
to get
. Recall by Pascal's triangle that
,
, so the leftmost digit is guaranteed to be
. Now, multiplying by our scale factor the answer is
~Geometry285