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.
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
Sometimes it is advantageous to express a value in round numbers. To round to a particular place, look at the digit immediately to the right of the one you want to round to. If that digit is 4, 3, 2, 1 or 0, zero it and everything to the right of it out. If that digit is 5, 6, 7, 8 or 9, increase your target digit by one and zero everything to the right of it out. Since the number to the right of the 1 is a 9, increase the 1 to 2 and zero everything else out. 199 rounds up to 200.
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.
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