2026/002) Let a,b,c be integers. If $4a+5b−3c$ is divisible by $19$, prove that $ 6a−2b+5c$ is also divisible by $19$

 As 4 and 6 have LCM 12 so we should multiply by 3 and proceed

$4a + 5b -3c$ is divisible by 19 so multiply by 3 to get $12a + 15b -9c$ is divisible by 19. adding $19c-19b$ we get

$12a - 4b +10c$ is divisible by 19

or $2*(6a-2b + 5c)$ is divisible by 19 and as 2 is co-prime to 19 hence $6a-2b+5c$  is divisible by 19

2026/001) How do I find the smallest positive integer n, such that each digit of n is either 0 or 7, and n is divisible by 15?

It is divisible by 15. So it is divisible by both 3 and 5

Because it is  divisible by 5 so unit digit is 0 or 5, As we have to chose from 0 and 7 so it is zero. Now we cannot have a zero in other position as it will increase the value of the number as it is additional digit and does not serve any other purpose.Sso other digits shall be 7 and we require 3 7s to be divisible by 3 or the number as 7770( 15 * 518) .

 

2025/034) If p can be expressed as sum of 2 squares say $p = x^2+y^2$ then express 2p ,5p, 19p as sum of 2 squares

If p can be expressed as sum of 2 squares as below

$p =  x^2+y^2$

We know $2= 1^2 + 1^2$

Now using the identity 

$(a^2+b^2)(x^2 + y^2) = (ax+by)^2 + (bx-ay)^2$ (we can expand and see the result

We get $2p = (x+y)^2+ (x-y)^2$

Further $5 = 2^2 + 1^2$

So we get $5p = (2x+y)^2 + (x-2y)^2$

But 19 is of the form 4n + 3 and it cannot be expressed as sum of 2 squares so 19p cannot be expressed as sum of 2 squares  

 

 

2025/033) Let a,b,c be the consecutive sandwiched integers between any cousin primes. Is it true that either $5∣(a+b+c)$ or $5| (a^2+b^2+c^2)$ ?

 Let x,y be cousin primes. We have x,a,b,c,y are 5 numbers.

now a = b-1 and c = b +1

so We have $a+b+c = 3b$

$a^2+b^2+c^2 = (b-1)^2 + b^2 + (b+1)^2 = 3b^2 + 2$

x is a prime so x cannot be multiple of 5 or b cannot be of the form 5k+2

y is a prime so y cannot be multiple of 5 or b cannot be of the form 5k +3

If b is of the form 5k $a+b+c$ is divisible by 5

If b is of the form 5k+ 1  $a^2+b^2+c^2 = 3(5k+1)^2 + 2 = 3(25k^2+10k+1) + 2 = 75k^2+ 30k + 5 = 5(15k^2+ 6k+1)$ multiple of 5

 If b is of the form 5k+ 4  $a^2+b^2+c^2 = 3(5k+4)^2 + 2 = 3(25k^2+40k+16) + 2$

$= 75k^2+ 120k + 50 = 5(15k^2+24k+10)$ multiple of 5

 So either of them is multiple of 5 

 

  

2025/032) How do I find the roots of the equation $x^3-14x^2+56x–64=0$, the roots are in geometric progression?

 roots are in GP so sw have root are $r, \frac{r}{a},ra$ so we get product of roots 

=$r^3= 64$ or $r=4$ So one root is 4 and hence (x-4) is a factor.  

we have by synthetic division.

$x^3-14x^2+56x–64=0= (x-4)(x^2-10x + 16)$

one factor is (x-4) and other factor is quadratic  and can be factored as $(^2-10x+16=(x-2)(x-8)$

giving roots 2,4,8 and they are in GP(just for cross checking) 

2025/031) Find a composite number for each of the following numbers to make each a pair of coprime numbers? 80, 63 ,135,88

 Let us factorize each of them

$80 = 2^4 * 5$

$63 = 3^2 *7$

$135 = 3^3 *5$

$88 = 2 ^ 3 * 11$

The prime factors are 2,3,5,7,11

 Any number  which is a composite number  and does not have factor any of above is the number and smallest is $13^2 = 169$

 

2025/030) Show that $12| (a-b)(b-c)(c-a)(a-d)(b-d)(c-d)$ for integers a,b,c,d.

Because of  symmetry a,b,c,d are interchangeable.

 For it to be a multiple of 12 we need to show that it is a multiple of 3 and 4 . This is so because 3 and 4 are co-primes.

 Let us show that it is multiple of 3. As there are 4 numbers a,b,c,d all 4 numbers divided by 3 cannot have 4 differences, So two of them as a and b have same remainder. then (a-b) i divisible by 3

Now let us show that it is is multiple of 4

If at least 3 of them say a,b,c are  even or odd then (a-b) and (b-c) both are even so multiple if product of 4.

If two say a and b are even and other 2 c and d are odd then (a-b) and (c-d) both are even and product is multiple of 4.

As product is multiple of 4 and 3 so it is multiple of 12.