To find the Highest Common Factor (HCF) of 52 and 117 and express it in the form 52x + 117y, we can use the Euclidean algorithm.

1. Step 1: Find the Remainder

   Divide the larger number by the smaller number and find the remainder.

   117=2×52+13117=2×52+13

   So, the remainder is 13.

2. Step 2: Replace Numbers

   Now, replace the divisor with the previous remainder, and the dividend with the divisor.

   52=4×13+052=4×13+0

   Here, the remainder is 0, so we stop. The divisor at this step, which is 13, is the HCF.

3. Expressing in the given form

   Now, we backtrack to express the HCF, which is 13, in the form 52x + 117y.

   Using the reverse steps of the Euclidean algorithm, we can express the HCF as a linear combination of 52 and 117:

   13=117−2×5213=117−2×52

   Hence, the HCF of 52 and 117 expressed in the form 52x + 117y is 13=117−2×5213=117−2×52.

To prove that x2−xx2−x is divisible by 2 for all positive integer xx, we can use mathematical induction.

Base Case: Let's start by checking the base case. When x=1x=1, x2−x=12−1=0x2−x=12−1=0. Since 0 is divisible by 2, the base case holds.

Inductive Hypothesis: Assume that x2−xx2−x is divisible by 2 for some positive integer kk, i.e., k2−kk2−k is divisible by 2.

Inductive Step: We need to show that if the statement holds for kk, then it also holds for k+1k+1.

(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k

By the inductive hypothesis, we know that k2−kk2−k is divisible by 2. And we know that kk is a positive integer, so kk is also divisible by 2 or it is an odd number.

• If kk is divisible by 2, then k2−kk2−k is divisible by 2, and kk is divisible by 2, so their sum (k2−k)+k(k2−k)+k is also divisible by 2.

• If kk is an odd number, then k2−kk2−k is still divisible by 2, and when you add kk to it, you still get an even number, which is divisible by 2.

In both cases, (k+1)2−(k+1)(k+1)2−(k+1) is divisible by 2.

Therefore, by mathematical induction, we conclude that x2−xx2−x is divisible by 2 for all positive integers xx.

Sure, let's prove this statement.

Let's start with m=2k+1m=2k+1 and n=2l+1n=2l+1, where kk and ll are integers.

Now, let's square both mm and nn:

m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1 n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1

Now, let's sum them:

m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2

Since 2k2+2k+2l2+2l2k2+2k+2l2+2l is an integer, let's denote it as qq. Then:

m2+n2=2q+2m2+n2=2q+2

This clearly shows that m2+n2m2+n2 is even, as it is divisible by 22.

To prove that m2+n2m2+n2 is not divisible by 44, let's consider the possible remainders when dividing by 44:

• If kk and ll are both even, then m2+n2m2+n2 will leave a remainder of 22 when divided by 44.
• If kk and ll are both odd, then m2+n2m2+n2 will also leave a remainder of 22 when divided by 44.

Thus, m2+n2m2+n2 is even but not divisible by 44 when both mm and nn are odd positive integers.

To find aa, we can use the relationship between the highest common factor (HCF) and the least common multiple (LCM).

Given: HCF(6,a)=2HCF(6,a)=2 LCM(6,a)=60LCM(6,a)=60

We know that the product of the HCF and LCM of two numbers is equal to the product of the numbers themselves. So, we have:

HCF(6,a)×LCM(6,a)=6×aHCF(6,a)×LCM(6,a)=6×a

Substituting the given values:

2×60=6×a2×60=6×a 120=6a120=6a

Now, we solve for aa:

a=1206a=6120 a=20a=20

Therefore, a=20a=20.

To find the smallest number that is divisible by both 90 and 144 when increased by 20, we need to find the least common multiple (LCM) of 90 and 144. Then, we'll add 20 to that LCM to get our answer.

First, let's find the LCM of 90 and 144.

The prime factorization of 90 is 2×32×52×32×5.

The prime factorization of 144 is 24×3224×32.

To find the LCM, we take the highest power of each prime factor that appears in either number:

LCM=24×32×5=720LCM=24×32×5=720

Now, we add 20 to 720:

720+20=740720+20=740

So, the smallest number that, when increased by 20, is exactly divisible by both 90 and 144 is 740.