The prime factorisation of 26 and 91 is
26 = 2 x 13
91 = 7 x 13
Hence, HCF (26, 91) = 13
If we solve, (3 + $\sqrt{7}$) (3 – $\sqrt{7}$), we get
(3 + $\sqrt{7}$) (3 – $\sqrt{7}$)
= $3^{2}$– $(\sqrt{7})^{2}$
= 9 – 7
= 2 [By $a^{2}$– $b^{2}$ = (a – b) (a + b)]
70 – 5 = 65 and 125 – 8 = 117
HCF (65, 117) is the largest number that divides 70 and 125 and leaves remainder 5 and 8.
HCF (65, 117) = 13
The least number will be LCM of 1, 2, 3, 4, 5.
Hence, LCM (1, 2, 3, 4, 5) = 2 x 2 x 3 x 5 = 60
$\dfrac{23}{(2^{2} . 5)}$ = $\dfrac{(23 × 5)}{(2^{2})}$
$5^{2}$ = $\dfrac{115}{(10)^{2}}$ = $\dfrac{115}{100}$ = 1.15
Hence, $\dfrac{23}{(2^{2} . 5)}$ will terminate after two decimal places.
Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.
396 > 231
Using Euclid’s division algorithm,
396 = 231 × 1 + 165
231 = 165 × 1 + 66
165 = 66 × 2 + 33
66 = 33 × 2 + 0
117 > 65
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
Therefore, HCF(65, 117) = 13
According to the given,
65m – 117 = 13
65m = 117 + 13
65m = 130
m = 130/65 = 2
We know that an odd number in the form (2Q + 1) where Q is a natural number ,
so, $n^{2}$ -1 = (2Q + 1)$^{2}$ -1
= 4Q$^{2}$ + 4Q + 1 -1
= 4Q$^{2}$ + 4Q
Substituting Q = 1, 2,…
When Q = 1,
4Q$^{2}$+ 4Q = 4(1)$^{2}$ + 4(1) = 4 + 4 = 8 , it is divisible by 8.
When Q = 2,
4Q$^{2}$ + 4Q = 4(2)$^{2}$ + 4(2) =16 + 8 = 24, it is also divisible by 8 .
When Q = 3,
4Q$^{2}$ + 4Q = 4(3)$^{2}$ + 4(3) = 36 + 12 = 48 , divisible by 8
It is concluded that 4Q$^{2}$ + 4Q is divisible by 8 for all natural numbers.
Hence, n$^{2}$ -1 is divisible by 8 for all odd values of n.
According to Euclid’s division lemma,
a = 3q + r, where 0 ≤ r < 3 and r is an integer.
Therefore, the values of r can be 0, 1 or 2.