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Q: What is a common multiple of 3 and 4 digit total 9 less than fifty?

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56

36

Yes. Digit total is 9 which is a multiple of 3 so it is too.

I'll guess those are three-digit numbers. The way to find if a number is a multiple of 3 is to total its digits. If that total is a multiple of 3, the whole number is a multiple of 3. 285 and 126, yes. 770, 176, 410, 452, 650, no.

There are multiple answers. The only criteria is that the last digit be one third of the second to last ( the tens digit ). So, 731 works as does 362 and 1,631, 2,162 and 11,162. You can also do 111,111,131.

48

Easier than what? The easiest way to produce a three-digit number that is a multiple of 3 (has 3 as a factor) is to make sure the digits total to a multiple of 3. 123, 555, 498 are all multiples of 3.

The LCM that satisfies the condition is 48.

2010Answer 2If you half it and you get a multiple of 3, then what you are saying is it is a multiple of 6 (2 x 3 = 6).So you want a 4 digit number, a number between 1000 and 9999, which is a multiple of 6.The lowest is 1002 (= 6 x 167), followed by 1008, 1014, 1020 ... etc. up to 9996 (= 6 x 1666).So, there are a total of 1500 such numbers.

Starting with the number just before the "decimal/binary" point and going left: you leave the first digit as it is, add 2 times the digit to its left, add 4 times the digit to its left, add 8 times the digit to its left, and so on to the end. The multiple is doubled each time. If you have digits to the right of the point, you add 1/2 times the first digit add 1/4 times the digit to its right, add 1/4 times the digit to its right, and so on to the end. The multiple is halved each time. The grand total is the equivalent decimal number.

It does not do anything!

With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.

900,000.

4

It is: 1+2 = 3

The first 8 digit number is 10,000,000, the last is 99,999,999 which means there are 90,000,000 8 digit numbers

the squares of 10 though 22 are all three digit numbers under 500. though none of the sums of their digits total an even multiple of 15.

in sigle digit -1 in double digit - 18 in triple digit - 19+(100+19) +19+19 total=195

All numbers divisible by 5 (of which 15 is a multiple) have a final digit of 0 or 5. All numbers divisible by 3 (of which 15 is a multiple) have the sum of the digits totalling 3 or a multiple of 3. Therefore, a number is divisible by 15 if the sum of its digits total 3 or a multiple of 3 and its final digit is 0 or 5. Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by 3. Final digit 5. This number is divisible by 15. (32085 ÷ 15 = 2139) 7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by 15.

nothing

Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991

-3

the right answer is 10 cause there is o,1,2,3,4,5,6,7,8,9 so there 10 numbers in total which have 1 digit

16, 25, 34, 43, 52, 61, 70 Seven of them.

You have 2 options for the first digit, 2 options for the second digit, etc. ... In total, that gives you 210 combinations.