Monday, January 23, 2012

Puzzle 26


Consider the three digit numbers 301, 431, 602, 715, 842, 856, 973, 986
They ALL satisfy the following properties:
1)  All  the digits are different
2)  When the digits are placed along a cirlce, each digit is three times the previous digit, or one unit more, or two units more (while doing this, keep in mind that if this number comes out to be more than 9, only the units digit is taken).  Hence 7 is followed by 1, 2 or 3.  3 is followed by 9, 0 or 1.  If the number ends with the digit 8, then the first digit can be 4, 5 or 6.
Not only the above listed numbers, their cyclic permutations also satisfy the above properties.
Hence the above list gives 22 three digit numbers.  We know, no number starts with a zero.
Now, try for four digited numbers that satisfy the above properties.  How many such numbers are there, if we consider their cyclic permutations also as different number?

Sunday, January 22, 2012

Puzzle 25 - Remainders and quotients


Start with N, a six digit number.  Subtract three from it, and then the new number formed should be divisible by 7.  Take the six sevenths of this number.  Call this new number i.e, as N1, and it should be divisible by 7.  Continue this process till you form N6.  Find N and N6.  How long you can go like this?

Friday, January 13, 2012

Puzzle 24 Trigonometry

What is the minimum of the absolute value of the sum of all the six trigonometric functions?

Thursday, January 12, 2012

Puzzle 23 - Polynomials

If the equation with real coefficients has both roots greater than 1, then show that the equation has at least two real roots.

Saturday, January 07, 2012

Puzzle 22 - Cresents on sides of right triangle


 
ABC is a triangle right angled at C.  Three semicircles are drawn on the sides AB, BC and CA as the respective diameters.  If the sides a, b, c of the triangle are positive integers and the sum of the shaded areas is also to be a positive integer, how many distinct such triangles are there with perimeter at most 60 units?

Thursday, January 05, 2012

Puzzle 21 - Something wrong with the question

Hi All
Following is the problem given in one of the Math Competitions.  I feel that there is some issue with the data.  Here is the problem.
" satisfying and

for all "

I have one function that satisfies the last two conditions but I have issue with the first condition.  Please go through the question and let me know your take.


Divisibility test - for 7

If is an ‘n’ digited number, it will be divisible by 7 if is divisible by 7.

Proof:
If , then the number represents the (n-1) digited number .

So, if is divisible by 7, then s is divisible by 7 – which can be easily verified from the relation that .

If the number resulted at this step is also high, the procedure can be continued till it can be conveniently decided whether the number is divisible by 7 or not.

Let us take the number 514199 which is actually equal to 7*73457, hence is divisible.
We check the above rule with this number
Step 1: 51419-18=51401
Step 2: 5140-2=5138
Step 3: 513-16=497 which is divisible by 7. Hence the number 514199 is divisible by 7.


Let us try with the number 479833024705.  This is NOT divisible by 7
Step 1: 47983302470-10=47983302460
Step 2: 4798330246-0=4798330246
Step 3: 479833024-12=479833012
Step 4: 47983301-4=47983297
Step 5: 4798329-14=4798315
Step 6: 479831-10=479821
Step 7: 47982-2=47980
Step 8: 4798-0=4798
Step 9: 479-16=463 which is not divisible by 7, hence the number 479833024705 is not divisible by 7

I will update this post with algebraic version to these calculations which makes the process simpler. 


Tuesday, January 03, 2012

Puzzle 20 - Three circles

are three circles.
Centre of lies on .
touch internally.
and cut at A and B. AB is extended to cut at X and Y.
and touch at M.
M is joined to X and Y. MX and MY cut at P and Q.
Prove that PQ is a tangent to .












I think the solution I have for this problem is little long. I am looking for simpler solution.

Monday, January 02, 2012

Puzzle 19 - Arrangement of Natural Numbers

All Natural numbers are arranged in the following pattern



Column:        1        2       3       4         5        6      7         8
Row 1           1        2       3                 13      14     15
Row 2                             4                 12               16
Row 3                             5                 11               17
Row 4                             6                 10               18
Row 5                             7        8       9                 19       20

If the pattern continues, in which Column the number 2012 appears?

Sunday, January 01, 2012

22nd Anniversary Brahmotsavas at our SARASWATHY TEMPLE

అందరికీ నూతన ఆంగ్ల సంవత్సరాది శుభాకాంక్షలు
Following is the information about the 22nd Anniversary Brahmotsavas at our SARASWATHY TEMPLE