Math IOQM

Q1. Consider the equation ab + bc + ca = abc - a - b - c. Is it necessary for all of a,b,c to be even? Solution: Yes. Consider the cases one by one when it is not so. Case 1: All odd. Rewrite the equation like this: a(b+1) + b(c+1) + c(a+1) = abc RHS = odd LHS = even + even + even = even Not possible. Case 2: a,b odd. c even. WLOG Rewrite the equation like this: a(b+1) + b(c+1) + c(a+1) = abc RHS = even LHS = even + odd + even = odd Not possible. Case 3: a odd. b,c even. WLOG Rewrite the equation like this: a(b+1) + b(c+1) + c(a+1) = abc RHS = even LHS = odd + even + even = odd Not possible. So we have checked all cases: 1 odd, 2 odd, 3 odd. None of them works. Only option is all even. Q2. Sum of all divisors of a natural number? Q3. If last 2 digits of 107n and n are same then what is the smallest value of n? Solution: 107n = n mod 100 => 106n = 0 mod 100 => 6n = 0 mod 100 => 6n = 100k => 3n = 50k => n has to be a multiple of 50 since 3 is not. So smallest n = 50. Q4....

Q. Find the sum of all non-negative integers a a such that the quadratic equation a 2 x 2 − ( 3 a 2 − 8 a ) x + 2 a 2 − 13 a + 15 a^2 x^2 - (3a^2 - 8a)x + 2a^2 - 13a + 15 has at least one integer root . Solution: Solution sketch (step-by-step) Step Reasoning 1 . Write the equation a 2 x 2 − ( 3 a 2 − 8 a ) x + ( 2 a 2 − 13 a + 15 ) = 0 a^{2}x^{2}-\bigl(3a^{2}-8a\bigr)x+\bigl(2a^{2}-13a+15\bigr)=0 . 2 . Compute the discriminant D D D = ( 3 a 2 − 8 a ) 2 − 4 a 2 ( 2 a 2 − 13 a + 15 ) D=\bigl(3a^{2}-8a\bigr)^{2}-4a^{2}\bigl(2a^{2}-13a+15\bigr) . 3 . Simplify D D D = a 4 + 4 a 3 + 4 a 2 = a 2 ( a + 2 ) 2 = [ a ( a + 2 ) ] 2 D=a^{4}+4a^{3}+4a^{2}=a^{2}(a+2)^{2}=\bigl[a(a+2)\bigr]^{2} – already a perfect square, so roots are always rational. 4 . Write the roots explicitly x = 3 a 2 − 8 a    ±    a ( a + 2 ) 2 a 2 x=\dfrac{3a^{2}-8a\;\pm\;a(a+2)}{2a^{2}} . Splitting the ± gives x 1 = 2 − 3 a , x 2 = 1 − 5 a \displaystyle x_{1}=2-\frac{3}{a},\qquad x_{2}=1-\frac...

Q. x 4 − q x 3 + 2 ( 10 − a ) x + 9 q x + a = 0 x^4 - qx^3 + 2(10-a)x - 9qx + a = 0 For x x where a a is real parameter, has no real roots for a ∈ ( − ∞ , − p q ) a \in (-\infty, -\tfrac{p}{q}) . p , q p, q are co-prime. Find p + q p+q . Solution: Divide by x^2: x 4 − 9 x 3 + 2 ( 10 − a ) x 2 + 9 a x + a 2 x 2 = 0 \frac{x^4 - 9x^3 + 2(10 - a)x^2 + 9ax + a^2}{x^2} = 0 x 2 − 9 x + 20 − 2 a + 9 a x + a 2 x 2 = 0 x^2 - 9x + 20 - 2a + \frac{9a}{x} + \frac{a^2}{x^2} = 0 ( x 2 + a 2 x 2 ) − 9 ( x − a x ) + 20 − 2 a = 0 \left(x^2 + \frac{a^2}{x^2}\right) - 9\left(x - \frac{a}{x}\right) + 20 - 2a = 0 Let x - a/x = k ⇒ x 2 + a 2 x 2 − 2 a x ⋅ x = k 2 \Rightarrow \quad x^2 + \frac{a^2}{x^2} - \frac{2a}{x} \cdot x = k^2 Simplifies to: x 2 + a 2 x 2 = k 2 + 2 a x^2 + \frac{a^2}{x^2} = k^2 + 2a Simplified and Substituted Expression: k 2 + 2 a − 9 k + 20 − 2 a = 0 k^2 + 2a - 9k + 20 - 2a = 0 Cancelling + 2 a +2a and − 2 a -2a : k 2 − 9 k + 20 = 0 k^2 - 9k + 20 = 0 Fa...

Q1. A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows are there? Solution: 1 + 3 + .. (2n-1) = 100 = n/2[2.1 + (n-1).2] => 100 = n.[1 + n - 1] = n^2 => n = 10.  Q2. The second and fourth terms of a geometric sequence are 2 and 6. Find the possible first terms. Solution: r^2 = 3, ar = 2 => a = -+2/sqrt(3) Q3.  Figures 0, 1, 2, and 3 consists of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many non-overlapping unit squares would there be in Figure 100? Solution: a1 = 1 + 4 a2 = 1 + 4 + 8 a3 = 1 + 4 + 8 + 12 a100 = 1 + (4 + 8 + 12 .. 100 times) = 1 + 100/2[8 + 99.4] = 1 + 200.[2 + 99] = 1 + 101.200 = 20201 Q4.  Let 1, 4, . . . and 9, 16, . . . be two arithmetic sequences. The set S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S? Solution: Total terms ...