Difference between revisions of "2005 AMC 10B Problems/Problem 20"

Revision as of 22:05, 31 May 2021

Problem

What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?

$\mathrm{(A)}48000\qquad\mathrm{(B)}49999.5\qquad\mathrm{(C)}53332.8\qquad\mathrm{(D)}55555\qquad\mathrm{(E)}56432.8$

Solution 1

We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 \div 120$ to get $\boxed{\mathrm{(C)}53332.8}$

Solution 2

We can first solve for the mean for the digits 1, 3, 5, 7, and 9 since each is 2 away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the 8 with a 9. We can start how be calculating the gain when the 8 was in the ones digit. Since there are $4! = 24$ numbers with the 8 in the ones digit and 1 was gain from each of them, 24 is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the 9 instead of the 8 to get $6666600-266664=6399936$ . Finally, we divide this by 120 to get the average. $6399936/120= \boxed{\mathrm{(C)}53332.8}$

Solution 3

The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$ . Values occur in every place so $4.8 * 11111 = 53332.8$ .

Revision as of 10:34, 24 April 2021 (view source) Naren pr (talk \| contribs) (→‎Solution 3) ← Older edit		Revision as of 22:05, 31 May 2021 (view source) Mobius247 (talk \| contribs) (→‎Problem) Newer edit →
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	== Problem ==		== Problem ==
−	What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 ~~and~~ exactly once?	+	What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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