50 and 51
To figure something like that out, subtract 1 and divide by 2 to get the smaller of the two.
101-1=100
100/2=50
Simple as that.
The difference between the squares of two consecutive integers j and j+1 is |2j+1|. There are therefore two such pairs where this quantity is 17:-9 and -88 and 9
Positive signed numbers with have a + Positive integers will not.
8
62 and 63
15 and 16
2 & 4, 28 & 30 etc. Any two even integers which differ by 2 in value.
I assume you mean what are the two consecutive integers. Algebraically; X = integers.X + (X + 1) = 2752X + 1 = 2752X = 274X = 137============solution setorSince the consecutive integers differ by 1, then the first number is 274/2 = 137, so the second one is 1 more, 138.
Consecutive prime numbers are 2 integers that differ by 1 and are both prime. Since 2 is the only even prime, 2 and 3 are the only consecutive primes.
Let x be the smallest of the consecutive odd integers. Since consecutive odd integers differ by 2, we havex + (x + 2) + (x + 4) + (x + 6) = -2204x + 12 = -220 (subtract 12 to both sides)4x = -232 (divide by 4 to both sides)x = -58Thus, the four consecutive odd integers whose sum is -220 are -58, -56, -54, and -52.
Since the numbers differ by 2 and the GCF is 2, they are consecutive even integers. So the numbers are 6 and 8.
24 and 25, which are (49-1)/2 and (49+1)/2
Let the first integer be x. Since the consecutive integers differ by 1, the other consecutive integers would be (x + 1), (x + 2), and (x + 3). So we havex + x + 1 + x + 2 + x + 3 = 904x + 6 = 904x + 6 - 6 = 90 - 64x = 844x/4 = 84/4x = 21Thus, the third integer is 23 (21 + 2 = 23).