What are all of the multiples of 14?
The multiples of 14 are 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, etc.
What 3 digits add up to 14?
If the digits in a 3-digit integer must add to 14, the answer is 70. And if a leading 0 is allowed, the answer is 75. Please try to be less ambiguous on future questions.
What is the sum of all 3 digit natural numbers?
Hence, the sum of all three-digit natural numbers, which are multiples of 7 is 70336. Note: Another formula for the sum of n terms is given by S=n2[a+l] where l is the last term of the series. This formula can be easily used in questions like these where we know the last term of the sequence.
What are the multiples of 3 and 14?
Step 1: List a few multiples of 3 (3, 6, 9, 12, 15, . . . ) and 14 (14, 28, 42, 56, 70, 84, . . . . ) Step 2: The common multiples from the multiples of 3 and 14 are 42, 84, . . . Step 3: The smallest common multiple of 3 and 14 is 42.
What are the pairs of 14?
The pair factors of 14 are (1, 14), (2, 7).
What are all 3 digit multiples of 14?
Between 100 to 1000 are 900 numbers. 900/14 =~ 64 so there will be 64 multiples of 14. The go from 112 = 8*14 to 71*14 = 994. The sum of these numbers are 14 (8 + ….. + 71). The sum 8 + …. + 71 = 79*64/2 = 2528. So the sum of all 3 digit multiples of 14 is 14*2528 = 35392.
What is the sum of all multiple of 3 from 1-1000?
I put the number 3 in the first cell (A1) of the spreadsheet program Excel. I then set cell A2 as A1+3. Then I copied the formula to all cells A3 to A333 and then set the formula in A334 as Sum (A1:A333) 3,6,9,12,……,999. n=332. The sum of all multiple of 3 from 1 to 1000 is equal to 3 times the sum of all natural numbers from 1 to 333.
How many 3-digit numbers are divisible by 21?
The first 3-digit number divisible by 14 is 112 (8 times), and the last 3-digit number is 994 (71 times). Therefore, there are 64 3-digit numbers divisible by 14, the summation of which is equal to 35392. The first 3-digit number divisible by 21 is 105 and the last 3-digit number is 987. Therefore, there are 43 3-digit numbers divisible by 21.
What is the sum of numbers from 100 to 999 divided by 3?
To solve your problem we need to add all the numbers from 100 to 999 and then subtract the numbers that can be divided exactly by 3. Now we need to calculate the sum of the first ⌈ 999 − 100 3 ⌉ = 300 terms of the sequence with a = 102 and d = 2, that is 300 2 ( 2 ( 102) + 3 ( 299)) = 165150, the result is then 494550 − 165150 = 329400
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