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How many strings of three decimal digits begin with an odd digit?
There are 500 of 3-digit numbers begin with an odd one (100-199, 300-301, 500-599,700-799, and 900-999). In terms of "how many strings of", may filter or expand the above answer.
When the largest two digit even number is added to the smallest three digit odd number what is the sum?
199
How many three-digit numbers are there in which the tens digit is greater than the ones digit?
Considering only positive integers then:We are looking at all the numbers between 100 to 999 (inclusive).Or 9 sets of 100.For each set of 100 (e.g. 100 to 199 or 200 to 299)There will be 0+1+2+3+4+5+6+7+8+9 = 45 numbers where this statement holds true.We know this because for 100 to 109 there are none.For 110 to 119 there is 1 (110)For 120 to 129 there are 2 (120 & 121)and this pattern holds all the way to 190 to 199 where there are 9.This will be true for each set of 100 numbers so the answer is 45 * 9 = 405.
When writing numbers from 1 to 10000 how many times is the digit 9 written?
To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
Why do we have 450 odd three digit numbers?
There are 899 three digit numbers (100 through 999).Each century (grouping of 100 numbers) has 50 odd numbers. (half of the numbers are odd, half are even)So.. 100-199, 200-299, 300-399, 400-499, 500-599, 600-699, 700-799, 800-899, 900-999We have 9 centuries.Number of odd numbers = 9*50 = 450
How many strings of three decimal digits begin with an odd digit?
There are 500 of 3-digit numbers begin with an odd one (100-199, 300-301, 500-599,700-799, and 900-999). In terms of "how many strings of", may filter or expand the above answer.
How many times does 3 appear in 1 to 1000?
Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9. Case 1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits You have 199 = 81 such numbers. However, 3 could appear as the first or the second or the third digit.
How many odd numbers are between 100 and 199?
50
How many digits are there in the numbers 100 - 199?
-99
How many 3 digit numbers have exactly 2 digits that are the same?
We have three cases: 1- Hundreds digit= Tens digit: In every hundred 9 numbers are in this pattern: 110-112-113-114-115-116-117-118-119 and we have 9 hundreds, so in this case there are 81 numbers. 2- Hundreds digit= Units digit: In every hundred 9 numbers are in this pattern: 101-121-131-141-151-161-171-181-191 and we have 9 hundreds, so in this case there are 81 numbers! 3- Tens digit= Units digit: In every hundred 9 digits are in this pattern: 100-122-133-144-155-166-177-188-199, and we have 9 hundreds, so in this case there are 81 numbers! Total = 81 + 81 + 81 = 243
199 plus 99 in real numbers?
It is: 199+99 = 298
What are the composite numbers to 199?
199 is a prime number; it has no composite factors.
When the largest two digit even number is added to the smallest three digit odd number what is the sum?
199
How many three-digit numbers are there in which the tens digit is greater than the ones digit?
Considering only positive integers then:We are looking at all the numbers between 100 to 999 (inclusive).Or 9 sets of 100.For each set of 100 (e.g. 100 to 199 or 200 to 299)There will be 0+1+2+3+4+5+6+7+8+9 = 45 numbers where this statement holds true.We know this because for 100 to 109 there are none.For 110 to 119 there is 1 (110)For 120 to 129 there are 2 (120 & 121)and this pattern holds all the way to 190 to 199 where there are 9.This will be true for each set of 100 numbers so the answer is 45 * 9 = 405.
Two prime numbers between 100 and 199 such that the tens digit is a prime number and the tens and ones digit taken together are a 2 digit prime number find the sum of these two prime numbers?
For the tens digit to be a prime number then it must equal 2, 3, 5 or 7. There are four 3-digit prime numbers that fit the above condition and also have the tens and units digits forming a 2-digit prime number. 131, 137, 173, 179. The person supplying the question may like to sum the various combinations.
When writing numbers from 1 to 10000 how many times is the digit 9 written?
To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
Why do we have 450 odd three digit numbers?
There are 899 three digit numbers (100 through 999).Each century (grouping of 100 numbers) has 50 odd numbers. (half of the numbers are odd, half are even)So.. 100-199, 200-299, 300-399, 400-499, 500-599, 600-699, 700-799, 800-899, 900-999We have 9 centuries.Number of odd numbers = 9*50 = 450
