Q8 - Integer equation practice problem

Question:
3x + 7y = 100
x,y are positive integers.
How many pairs of x,y are possible?

Solution:
Given that x,y >= 1.

So 7y < 100
y < 100/7
y <= 14
So 1 <= y <= 14.
Now 3x = 100 - 7y
If you put all the values of y one by one you see that RHS becomes divisible by 3 for y = 1,4,7,10,13. So answer is 5 such pairs are there.

But why did we choose y for iterating and not x?
Let's do it with x.
3x < 100 => x <= 33
So we would have to iterate 33 times and check for divisibility by 7 each time. So it is wiser to go for y.

We can make the solution much simpler using modulo arithmetic.

3x = 100 - 7y
Do modulo by 3 on both sides.
3x % 3 = (100 - 7y) %3
0 = 100 % 3 - 7y %3   %3
0 = 1 - (7 %3) (y%3) %3
0 = 1- 1.y % 3
y = 1 % 3

Since y goes from 1 to 14, there are 5 values which leave the remainder 1 when divided by 3.

Comments

Post a Comment

Popular posts from this blog

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more

Combinatorics DPP - RACE 6 - Q16 pending discussion

Image
Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...
Read more

Algebra DPP 1.3 Quadratics

 Q1 . For what values of b do the equations: 1988x² + bx + 8891 = 0 and 8891x² + bx + 1988 = 0 have a common root? Solution: We are given two quadratic equations: 1988 x 2 + b x + 8891 = 0 1988x^2 + bx + 8891 = 0 8891 x 2 + b x + 1988 = 0 8891x^2 + bx + 1988 = 0 We are to find the values of b b for which they have a common root . Step 1: Let α \alpha be the common root. Then α \alpha satisfies both equations. So, From equation (1): 1988 α 2 + b α + 8891 = 0 (i) 1988\alpha^2 + b\alpha + 8891 = 0 \quad \text{(i)} From equation (2): 8891 α 2 + b α + 1988 = 0 (ii) 8891\alpha^2 + b\alpha + 1988 = 0 \quad \text{(ii)} Step 2: Subtract equation (i) from equation (ii): ( 8891 α 2 − 1988 α 2 ) + ( b α − b α ) + ( 1988 − 8891 ) = 0 (8891\alpha^2 - 1988\alpha^2) + (b\alpha - b\alpha) + (1988 - 8891) = 0 ( 8891 − 1988 ) α 2 − ( 8891 − 1988 ) = 0 (8891 - 1988)\alpha^2 - (8891 - 1988) = 0 6903 α 2 − 6903 = 0 6903\alpha^2 - 6903 = 0 α 2 = 1 ⇒ α = ± 1 \alpha^2 = 1 \Rightar...
Read more