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# Here's a fun puzzle!

A,B,C,D,E,F,G are 7 different integers.When properly arranged,they are consecutive integers lying between 30 and 39. 1) E - D + 4 = g/4

2)The highest number B Is a prime number.

3)C- G = B - A

4) A is not an even number. Find the numbers. Have fun fellas!

Note by Sridhar Thiagarajan
4 years, 7 months ago

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a=33,b=37,c=36,d=31,e=35,f=34,g=32

- 4 years, 7 months ago

Nice job!

- 4 years, 7 months ago

thanks

- 4 years, 7 months ago

4.Let X be the midpoint of the side AB of Δ ABC .Let Y be the midpoint of CX.Let BY cut AC at Z.Prove that AZ=2ZC.

- 4 years, 7 months ago

Much easier solution :

[AYX] = [BXY] = [CBY]

CZ/AZ = [CBZ]/[ABZ] = {[AZY] + [ABY]}/{[CZY] + [CBY] = 1/2

- 4 years, 7 months ago

nice job...........................

- 4 years, 7 months ago

Solution :

Let BW also be a median and let their intersection, the centroid, be G. Draw XW, BW, YW so we could compare area. We denote the area of a closed figure A by [A]. Let [XGB] = 1. Then, by the centroid theorem, [WGX] = 0.5, [WAX] = 1.5, [WGY] = 0.25, [BGY] = 0.5, [BCY] = 1.5. Let [WYZ] = a, [YZC] = b. Then

$\frac{a}{b} = \frac{0.25 + 0.5 + a}{b + 1.5}$ Cross-multiplying yields 2a = b, so $$a : b = 1 : 2$$. Since W is the midpoint of AC, AW : WZ : ZC = 3 : 1 : 2, or AZ : CZ = 2 : 1, or AZ = 2CZ.

- 4 years, 7 months ago

do this....An n-digit positive integer N is a Kaprekar number if the sum of the number formed by the last n digits in N2, and the number formed by the first n (or n−1) digits in N2 equals N. For example, 297 is a Kaprekar number since 2972 = 88209 and 88 + 209 = 297. There are five Kaprekar numbers < 100. Find them....and this There are four integers between 100 and 1000 that are each equal to the sum of the cubes of its digits. Three of them are 153, 371, and 407. Find the fourth number.

- 4 years, 7 months ago

do this

- 4 years, 7 months ago

I Don't like proofs :P

- 4 years, 7 months ago