The product in
Find the Last Two Digits has several multiples of 5. Erase them from the product to form a new product:
7 x 19 x 31 x 43 x 67 x 79 x 91 x 103 x 127 x ........ x 2023
Find the last two digits of this new expression.
A computer program is rather simple, so I request an evaluation using analytical techniques.
The last two digits are 41.
300/12=25. 25-5 multiples of 5=20. So the product has 20 numbers in every 300. The last two digits (sorted by last digit) are:
11,31,51,71,91
03,23,43,63,83
07,27,47,67,87
19,39,59,79,99
Almost all of these pair off to produce a number ending with 01. (ex. 11*91=1001, 3*67=20183*47=3901). The exceptions are 51 and 99, which pair with themselves: 51^2=2601, 99^2=9801.
The sequence ends as 2023 which is almost 6 cycles but leaves out 2047, 2059, 2071, 2083. This leaves 83,39,31,47 unpaired. 83 and 47 are a pair anyway so really just 39 and 31. (39*31 ends in 09)
That leaves 51^7 * 99^7 * 39 *31. But again we can remove 51 and 99 in pairs leaving one each. (51*99 ends in 49), so the final answer is the last two digits of:
09*49 which is 41
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Posted by Jer
on 2024-01-20 12:19:01 |
Solution
