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Q: How many three digit numbers contain at least two sevens?

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271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.

what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?

There aren't going to be any 7's in the hundreds digit, so first we look at how many there will be in the ones digit. Since there will be a 7 for every 10 numbers, and there are 501 numbers between 100 and 600 inclusive), we'll have floor(501 / 10) = 50 sevens in the ones digit. For the tens digit, we'll encounter 10 sevens for every 100, from 70 to 79, and so we have floor(501 / 100) = 5 times we'll have 70 - 79 so that's 5 * 10 = 50 sevens in the tens digit. However, we don't want to know how many sevens there are, we want to exclude overlaps. Thus we subtract every instance of 77 we see (overlap!), and we have floor(501 / 100) = 5 overlaps. 50 + 50 - 5 = 95, so a total of 95 numbers will have the digit 7 at least once.

Greatest: 989949 Least: 100000

We are to calculate the number of 1-digit to 6-digit numbers (i.e. numbers less than one million) that have at least one 1. Let's divide this into six cases by the number of digits in the number.For one digit numbers, the only number is 1 - 1 such number.For two digit numbers, we can find the number which do not contain a 1 more easily. There are 8 possible digits for the tens place (0 and 1 are not allowed), and 9 digits for the ones place (1 is not allowed). Thus we have 8 * 9 = 72 which do not work. There are 90 two digit numbers, so 90 - 72 = 18 such numbers.For three digit numbers, we will proceed in the same manner. There are 900 total 3 digit numbers, of which 8 * 9 * 9 = 648 numbers which do not contain a 1. The remaining 900 - 648 = 252 numbers contain at least 1 one - 252.For four digit numbers, continuing in the same manner we have 8 * 9 * 9 * 9 = 5832 which are not good out of the 9000 total. So, 9000 - 5832 = 3168.For five digit numbers, we have 8 * 9 * 9 * 9 * 9 = 52488. The remaining 90000 - 52488 = 37512 are good.For six digit numbers, we have 8 * 9 * 9 * 9 * 9 * 9 = 472392. The reamining 900000 - 472392 = 427608 are good.Finally, adding all of the totals, we have 1 + 18 + 252 + 3168 + 37512 + 427608 = 468,559 numbers between 1 and 999999 which contain at least 1 one.

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252

252

46000

271 of the first 1000 natural numbers contain at least one digit 5. That is 27.1 % of them.

i think alot like 100 at least

what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?

There aren't going to be any 7's in the hundreds digit, so first we look at how many there will be in the ones digit. Since there will be a 7 for every 10 numbers, and there are 501 numbers between 100 and 600 inclusive), we'll have floor(501 / 10) = 50 sevens in the ones digit. For the tens digit, we'll encounter 10 sevens for every 100, from 70 to 79, and so we have floor(501 / 100) = 5 times we'll have 70 - 79 so that's 5 * 10 = 50 sevens in the tens digit. However, we don't want to know how many sevens there are, we want to exclude overlaps. Thus we subtract every instance of 77 we see (overlap!), and we have floor(501 / 100) = 5 overlaps. 50 + 50 - 5 = 95, so a total of 95 numbers will have the digit 7 at least once.

100,000

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0

1009

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