The greatest three digit number that is divisible by 7 is 994.
There are 282 such numbers.
Best way to work this out: find the highest number below 10,000 that is divisible by 7 (9,996) and divide that by 7 (1,428). 1,428 is the amount of one-, two-, three- and four-digit numbers divisible by 7. Now find the number of one-, two- and three-digit numbers divisible by 7 (which is 994/7 = 142) and subtract this number from 1,428, giving us 1,286.
13
98 is the largest two digit number divisible by 7.
The greatest three digit number that is divisible by 7 is 994.
There are 282 such numbers.
Best way to work this out: find the highest number below 10,000 that is divisible by 7 (9,996) and divide that by 7 (1,428). 1,428 is the amount of one-, two-, three- and four-digit numbers divisible by 7. Now find the number of one-, two- and three-digit numbers divisible by 7 (which is 994/7 = 142) and subtract this number from 1,428, giving us 1,286.
13
98 is the largest two digit number divisible by 7.
The smallest 3-digit multiple of 7 is 105 = 15*7 The largest 3-digit multiple of 7 is 994 = 142*7 So there are 142-14 = 128 3-digit multiples of 7, ie 128 3-digit numbers that are divisible by 7.
Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.
9
48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.
For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.If the final digit of the number is 0 or 5, the original number is divisible by 5.Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.
1500000
Not all of them are.