Problems Of the Week (POW-1201s)

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The following addition is incorrect if considered in base-10. Find the base in which this addition is correct.

If we look closely the sum of unit digits and tens digits then we find that 6 + 7 + 5 + 8 = 6 + 8 + 8 + 4 = 26 in base-10. So in this new base also, they will be giving same result. And by addition of tens digits we are sure that they add up to a 2-digit number. So it can be easily concluded that 6 + 7 + 5 + 8 = 6 + 8 + 8 + 4 = 2610 = 12b, where b is the new base. Expanding the values in base-10, we get 26 = b + 2, i.e. b = 24.

How many numbers are there that appear both in the arithmetic sequence 4, 11, 18, 25, ... 2020 and the arithmetic sequence 4, 13, 22, 31, ..., 2020? The common numbers in the two sequence will be at a gap of LCM(7, 9) = 63. So we just need to find the number of terms in the arithmetic sequence: 4, 67, 130, , 2020. Number of terms will not change, if 59 is added to each of the terms. That means, we need to find the number of terms in the arithmetic sequence: 63, 126, 189, , 2079. i.e. number of terms in : 163, 263, 363, ., 3363 i.e. 33 numbers. Two equilateral triangles of side-length 4 are drawn on two adjacent sides of a square as shown. The distance between the two vertices A and B can be written as p + q. Find p q.

Let O be the center of the square then AO and BO both pass through the mid points of respective sides of squares and AB becomes the hypotenuse of the isosceles right triangle AOB such that AO = BO = 2 + 23. So AB = AO + BO = 2AO => AB = AO2 = (2 + 23)2 = 22 + 26 = 8 + 24. i.e. if p = 8, then q = 24 or vice-versa. But in both the cases p q = 8 24 = 192. Five unit circles are drawn such that every circle is touching only its two adjacent circles. Also a belt is wrapped around the circles as shown in figure. Find the length of the belt.

If we join the centers of the circle with point of contacts with belt wrapped, belt makes right angle with that. So the complete belt length is divided into five straight lines of length 2 units each and five circular arcs which, when combined, form circumference of a unit circle. Hence the required belt length = 25 + 2pi = 10 + 2pi. What is the sum of all natural numbers which are less than 10100 and are co-prime to it? It is simply 10100 (10100) =2 10199. [Remember that (N) is number of numbers less than N and co-prime to N and is given by (N) = N(1 - 1/a)(1 1/b) where a, b are the only distinct prime numbers contained in N]