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The trigonometry problem for IIT/Olympiad aspirants or anyone preparing for advance competition is reposted as none of the solution receeived is correct one. You can post your answer at www.facebook.com/iqiitjee.

What’s the minimum value of | sinx + cosx + tanx + cotx + secx + cosecx | ?

IQe Academy IQe provides intensive IIT/JEE math education to compete at IIT or an equivalent level anywhere in th

Here is a problem from trigonometry for IIT/Olympiad aspirants. Pl post your solution on facebook.com/iqiitjee.

What’s the minimum value of | sinx + cosx + tanx + cotx + secx + cosecx |

IQe Academy IQe provides intensive IIT/JEE math education to compete at IIT or an equivalent level anywhere in th

Here is the solution of 10 grader math problem for aspirants of IIT/Olympiads.
The problem: Find all natural numbers x such that the product of their digits (in decimal notation) is equal to x² - 10x – 22.
Solution:
Let the decimal expansion of x be d(1)d(2)d(3)….d(n), where d(i) are base-10 digits.
We, therefore, have x = d(1)*10^(n-1) + d(2)*10^(n-2) + ……+ d(n),
And that implies x ≥ d(1)*10^(n-1).

However, the product of the digits of x is
d(1)d(2)d(3)…d(n) ≤ d(1)*10*10…*10 = d(1)*10^(n-1).

The equality above holds only when x is a one-digit integer. Therefore the product of the digits of x is always at most x, with equality only when x is single digit.

That implies, x² - 10x – 22 ≤ x or x² - 11x – 22 ≤ 0.

It’s a simple inequation [ x(x – 11) – 22 ≤ 0]. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these numbers for the product of its digit.

Next, the number from 1 thru 11 are easily ruled out as their product (x² - 10x - 22) will produce negative values. However, for x = 12, we’ve, 12² - 10*12 – 22 = 2, which is the product of the digits of 12.

Therefore 12 is the only natural number with the desired properties.

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As promised, here is a class 10 (?) math problem for IIT/Olympiad aspirants that’s seriously gonna challenge your analytical skill!

Find all natural numbers x such that the product of their digits (in decimal notation) is equal to x² - 10x – 22.

Feel free to share with your friends/teachers. Answer will be posted by Sunday on facebook.com/iqiitjee.

IQe Academy IQe provides intensive IIT/JEE math education to compete at IIT or an equivalent level anywhere in th

The solution of the problem is outlined below.
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Substitute y = x + 1/x, in the re-arranged eq. as (x² + 1/x²) + a(x + 1/x) + b = 0, we get, y² + ay + b - 2 = 0.

and so we've, y = [- a ± √(a² + 8 – 4b)]/2

We also have |y| = |x + 1/x| ≥ 2. Therefore,

│(- a ± √(a² + 8 – 4b))/2│ ≥ 2, or [|a| + √(a² + 8 – 4b)]/2 ≥ 2 or
|a| + √(a² + 8 - 4b) ≥ 4

Rearranging and squaring both sides,
a² + 8 – 4b ≥ a² - 8|a| + 16 or 2|a| - b ≥ 2.

So a² + b² ≥ a² + (2 – 2|a|)² = 5a² - 8|a| + 4 = 5(|a| - 4/5)² + 4/5

Therefore the smallest possible value of a² + b² is 4/5 when a = ±4/5 and b = -2/5.

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Try this out from quadratic equation for students preparing for IIT/Olympiads. Post your solution on facebook.com/iqiitjee.

Let a and b be the real numbers for which the equation x^4 + ax^3 + bx^2 + ax + 1 = 0 has at least one real solution. For all such pairs (a, b), find the minimum value of a^2 + b^2.

IQe Academy IQe provides intensive IIT/JEE math education to compete at IIT or an equivalent level anywhere in th

Try this out from quadratic equation for students preparing for IIT/Olympiads.

Let a and b be the real numbers for which the equation x^4 + ax^3 + bx^2 + ax + 1 = 0 has at least one real solution. For all such pairs (a, b), find the minimum value of a^2 + b^2.

Here is the solution of the problem for math students preparing for advance competitions.
(((In a mathematical contest, three problems, A, B, and C were posed. Among the participants there were 25 students who solved at least one problem each. Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C. The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem. Of all students who solved just one problem, half did not solve problem A. How many students solved only problem B?))).

The generic approach may lead to solution in many such problems. Will post more advance sums of this kind latter.

Solution:
Let a, b, and c are no. of students who answered A, B, and C ONLY.
Let ab, bc and ca be the no. of students who answered ONLY two problems A-B, B-C, and C-A.
Let abc be the no. of students who answered all three A-B-C problems.

P.S. ab is not product of a and b. It represents students who solved both A and B etc.

We’ve to find b.

The condition, (Of all students who solved just one problem, half did not solve problem A), implies that,
(a+b+c)/2 = b+c => a = b+c ------------------------------- (1)

The condition, (The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem.), implies that,
a = ab + ca + abc + 1 ----------------------- (2)

The condition, (Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.), implies that,
b + bc = 2* (c + bc). => b = 2c + bc ----------------------- (3)

The condition (Among the participants there were 25 students who solved at least one problem each), implies that,
a + b + c + ab + bc + ca + abc = 25 ----------------------- (4)

There are 4 equations with 7 unknown variables! Using 1, 2, 3 in 4, we finally get,
4b + c = 26 ----------------------- (5)

Lets rewrite eq. 5 and explore possible solutions for the fact that all numbers are positive integers.

c = 26 – 4b ---------------------------- (6)

For b = 1, c = 22, a = 23. But this will not satisfy eq.4 (ab+bc+ca+abc = - 21), and hence not valid.
For b = 2, c = 18, a = 20. As above, this will not satisfy eq 4, and hence not valid.
For b = 3, c = 14, a = 17, As above, this will not satisfy eq. 4, and hence not valid.
For b = 4, c = 10, a = 14, As above, this will not satisfy eq. 4, and hence not valid.
For b = 5, c = 6, a = 11. This is also not valid as eq. 4 reduces to ab+bc+ca+abc = 3.
For b = 6, c = 2, a = 8, This is a valid solution as it doesn’t contradict any eqs above.
For b = 7 onward, c is –ve and no other solution is possible.

So b = 6.is a valid solution.

Here is a problem which only requires good analytical abiity – a prerequisite for IIT / Olympiads. Post your solution on facebook www.facebook.com/iqiitjee/.

In a mathematical contest, three problems, A, B, and C were posed. Among the participants there were 25 students who solved at least one problem each. Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C. The number of students who solved only problem A was one more than the number of students who solved A and at least one other problem. Of all students who solved just one problem, half did not solve problem A. How many students solved only problem B?

IQe Academy IQe provides intensive IIT/JEE math education to compete at IIT or an equivalent level anywhere in th

Here is the solution for those who have tried their heart out in solving it.

The problem:

“Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.”

The solution:

Let N = 100a + 10b +c for some digits a, b and c.

Then 100a +10b + c = 11m for some m.
We also have m = a² + b² + c².

Substituting this into the first equation and simplification, we get
100a + 10b + c = 11a² + 11b² + 11c².

For an integer divisible by 11, the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by 11. Thus we get: b = a + c or b = a + c -11.

Case 1: Let b = a + c.
We get, 100a + c +10a + 10c = 11a² +11c² + 11(a + c)² or 10a + c = 2a² + 2ac + 2c²

Since the right side is even, the left side must also be even. Let c = 2q for some q = 0,1,2,3,4.

Then 10a + 2q = 2a² + 4aq + 8q² or 5a + q = a² + 2aq + 4q².

Substitute q = 0,1,2,3,4 into the last equation and then solve for a.
When q = 0, we get a = 5. Thus c = 0 and b = 5. We get that N = 550, which works.
When q = 1, we get that a is not an integer. There is no N for this case.
When q = 2, we get that a is not an integer. There is no N for this case.
When q = 3, we get that a is not an integer. There is no N for this case.
When q = 4, we get that a is not an integer. There is no N for this case.

Case 2: Let b = a + c - 11. We get

110a + c + 10a + 10c - 110 = 11[a² + (a + c)² - 22(a + c) + c² + 121] or
10a + c = 2a² + 2c² + 2ac – 22a – 22c + 131 or
2(a - 8)² + 2(c – 23/4)² + 2ac – 505/8 = 0

Now we test all c = 0 thru 10. When c = 0, 1, 2, 4, 5, 6, 7, 8, 9, we get no integer solution to a. Thus, for these values of c, there is no valid N. However, when c = 3, we get
2(a – 8)² + 6a – 48 = 0.

We get that a = 8 is a valid solution. For this case, we get a = 8, b = 0, c = 3, so N = 803, and this is a valid value. Thus, the answers are N = 550, 803.

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