from 1 to 100 how many times 9 comes?
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9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99. You have not mentioned if only natural numbers must be considered while counting the number of 9s that appear between 1 and 100. If only natural numbers are considered while counting, then 9 appears 20 times.
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There are 10,000 possible combinations. Start at 0001, then 0002, then 0003...and work your way to 9999. Don't forget to try 0000 as well. You can also start at 9999 and work your way down to 0000 if you'd rather.
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9 is 45% of 20. If it is not obvious how to make the base 100 -- Look directly at the ratio of the Amount to the Base.
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1 Expert Answer 63 is 9 times as many as 7, and it is 7 times as many as 9.
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All numbers except the ones below are divisible by 2 or 3. 66 numbers between 1 and 100 inclusive are either divisible by 2 or 3. 1 is not include in the count. 99÷3=33 , 99÷2=49 .
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There are 24 numbers between 0 and 100 which are divisible by 4.
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Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
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7,17,27,37,47,57,67,70,71,72,73,74,75,76,77,78,79,87,97. the answer is 20!!!!!!! (77) is counted twice. Count!
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Answer Expert Verified The expression is not equivalent to other three is 9(n-9) .
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Multiples of 9 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ...
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50 percent (calculated percentage %) of what number equals 9? Answer: 18.
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Given 100, choose 98, there are 161,700 combinations. The number of combinations gets bigger fast. 100, choose 95, is over 7 million combinations. The largest number of combinations given a set of 100 elements, will be 100, choose 50.
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Since we're dealing with numbers between 1 and 100, we cannot include 1 and 100 in our calculations. So, there are 98 numbers between 1 and 100.
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Consider the following numbers which are divisible by 9, using the test of divisibility by 9: 99, 198, 171, 9990, 3411. Sum of the digits of 99 = 9 + 9 = 18, which is divisible by 9. Hence, 99 is divisible by 9.
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The number of 5-digit combinations is 10 5=100,000.
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So there are 81 numbers. However, 0 cannot come in both the places as we are excluding the number 100.. Hence, there are 80 ones. Let us fix two ones in unit and ten's places and hundred's place other than 1.
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So 10+10=20 times. That's the base ten answer. As a general rule, in base n, the digit 2 appears 2n times between 1 and 100, for all base n>2.