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.
199
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.
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.
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
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.
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.
50
200
-99
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
It is: 199+99 = 298
199 is a prime number; it has no composite factors.
199
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.
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.
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.