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I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I
Reqd no. is 9 -digits.. available digits are 1 2 3 4 5.. Numbers which are divisible by 4 can be determined by the last two digits.. from the given combinations,, the numbers div by 4 are 12, 24, 32, 44, 52 so.. the number's last two digits can be any of the above 5. Hence, we have to calculate the combinations for the first 7 digits.. Ans: (5C1)^7 * 5 ie 5^8 = 390625
3 digits numbers: from 100 to 999 The first number greater than 100 and divisible by 7 is 105 The last number lower than 999 and divisible by 7 is is 987 (987-105)/7=126 intervals of 7-length In fact 127 as numbers are to be counted, not intervals
Not evenly. An easy way to tell is that its digits add up to 7, a number which is not divisible by 3.
1- Any number is divisible by 1. 2- Any even number is divisible by 2. 3- If its digits add up to a number divisible by 3, the whole number is divisible by 3. 4- If the last two digits are divisible by 4, the whole number is divisible by 4. 5- If it ends in 5 or 0, the whole number is divisible by 5. 6- If it's even and also divisible by 3, the whole number is divisible by 6. 7- There is no rule for 7. 8- If the last three digits are divisible by 8, the whole number is divisible by 8. 9- If its digits add up to a number divisible by 9, the whole number is divisible by 9. 10- If it ends in 0, the whole number is divisible by 10.
Nope - in order for a number to be divisible by 9, the sum of its digits must also be divisible by 9. In this case, the digits 2+4+2+7=15.
If the number is even, it's divisible by 2. If the sum of the digits is a multiple of 3, the whole number is divisible by 3. If the last two digits are a multiple of 4, the whole number is divisible by 4. If the last digit is a 0 or a 5, the whole number is divisible by 5. If the number is even and divisible by 3, it's divisible by 6. If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7. If the last three digits are a multiple of 8, the whole number is divisible by 8. If the sum of the digits is a multiple of 9, the whole number is divisible by 9. If the number ends in 0, it's divisible by 10.
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.
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7800 is divisible by:2 - because it is 2 if it is even3 - the sum of its digits (7+8+0+0 = 15) is a multiple of 3 :4 - the last two digits is a multiple of 45 - it ends or 5 with 06 - it is divisible by 2 and 38 - the last three digits (800) is multiple of 810 - it ends with 0
No, because if you add the three digits together, you get 7 which isn't divisible by 3