41-- Less than 100-- 42 is the multiple of 7-- 4 + 1 = 5
five facts that involve positive numbers less than one-millionth.
9 is.
4
For a number to be a perfect square, its prime factors must be separable into two identical sets. For a number to be a perfect cube, its prime factors must be separable into three identical sets. For a number to be a perfect 'fifth', its prime factors must be separable into five identical sets. ---- If N x 2 is separable into two identical sets of prime factors, then: * N must have an odd number of 2's in its prime factorisation. * All other prime factors of N must occur an even number of times. If N x 3 is separable into three identical sets of prime factors, then: * The prime factorisation of N must contain a number of 3's that is one less than a multiple of three. * All other prime factors of N must occur a number of times that is already a multiple of three. If N x 5 is separable into five identical sets of prime factors, then: * The prime factorisation of N must contain a number of 5's that is one less than a multiple of five. * All other prime factors of N must occur a number of times that is already a multiple of five. ---- From the above, we know that N's prime factorisation must contain 2's, 3's and 5's. How many times does 2 appear in the prime factorisation of N? - It must be an odd number. - It must be a multiple of three. - It must be a multiple of five. The smallest valid frequency is 15. How many times does 3 appear in the prime factorisation of N? - It must be an even number. - It must be one less than a multiple of three. - It must be a multiple of five. The smallest valid frequency is 20. How many times does 5 appear in the prime factorisation of N? - It must be an even number. - It must be a multiple of three. - It must be one less than a multiple of five. The smallest valid frequency is 24. ---- Therefore N = 215 x 320 x 524 = 6810125783203125000000000000000. That seems to be the smallest possible value.
41-- Less than 100-- 42 is the multiple of 7-- 4 + 1 = 5
One
In that range, every prime number is either one more or one less than a multiple of 4. This also works for 6.
five facts that involve positive numbers less than one-millionth.
989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
144
144
9 is.
4
For a number to be a perfect square, its prime factors must be separable into two identical sets. For a number to be a perfect cube, its prime factors must be separable into three identical sets. For a number to be a perfect 'fifth', its prime factors must be separable into five identical sets. ---- If N x 2 is separable into two identical sets of prime factors, then: * N must have an odd number of 2's in its prime factorisation. * All other prime factors of N must occur an even number of times. If N x 3 is separable into three identical sets of prime factors, then: * The prime factorisation of N must contain a number of 3's that is one less than a multiple of three. * All other prime factors of N must occur a number of times that is already a multiple of three. If N x 5 is separable into five identical sets of prime factors, then: * The prime factorisation of N must contain a number of 5's that is one less than a multiple of five. * All other prime factors of N must occur a number of times that is already a multiple of five. ---- From the above, we know that N's prime factorisation must contain 2's, 3's and 5's. How many times does 2 appear in the prime factorisation of N? - It must be an odd number. - It must be a multiple of three. - It must be a multiple of five. The smallest valid frequency is 15. How many times does 3 appear in the prime factorisation of N? - It must be an even number. - It must be one less than a multiple of three. - It must be a multiple of five. The smallest valid frequency is 20. How many times does 5 appear in the prime factorisation of N? - It must be an even number. - It must be a multiple of three. - It must be one less than a multiple of five. The smallest valid frequency is 24. ---- Therefore N = 215 x 320 x 524 = 6810125783203125000000000000000. That seems to be the smallest possible value.
Every prime number except 2 must be odd. This is because if it were even it would be divisible by 2 and so wouldn't be a prime number. The next point to note is that all odd numbers are one more or one less than a multiple of 4. Because 2 more than a multiple of 4 is even, and 3 more than a multiple of 4 is the same as 1 less than the multiple of 4 above. Thus every prime number except 2 must be 1 more or 1 less than a multiple of 4.
No.3 is less than 5, thus cannot be a multiple of it; 3 is prime, therefore cannot be a multiple of any number besides 3 (which is one of its factors); by definition, 3 is ismply not a multiple of any number but 1 and 3.