Q.310. Useful property of multiplication.

Question 310.
What is the single-digit sum of the digits in 4444^4444 ?

Answer 310.
(4444)^2 = 4444 * 4444 will have sum = (4+4+4+4) x (4+4+4+4) = 4 in single digit
(4444)^3 = 4444 * (4444)^2 will have sum = 4 * (4+4+4+4) = 1 in single digit
(4444)^4 = 4444 * (4444)^3 will have sum = 1 * (4+4+4+4) = 7 in single digit
(4444)^5 = 4444 * (4444)^4 will have sum = 7 * (4+4+4+4) = 4 in single digit
Thus, with index 2, 3, 4, 5 the single digit sum is 4, 1, 7, 4
=> with index 2, 5, 8, 11, .... 2 + (n-1)*3 = 3n-1 the sum will be 4
=> sum will be
4 for index 4442,
1 for index 4443 and will be 
7 for index 4444.
Answer : 7.

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Q.192. Trigonometry puzzle

Question 192.
Two right triangles with common hypotenuse form quadrilateral that lying in the square with all 4 vertices on square's sides, as shown in the picture. Find the exact value of the length of the side of the square.

Answer 192.
Let the angle between side 7 and the vertical line be x
=> angle between side 24 and horizontal line = x
and angle between side 20 and vertical line
= 180° - x - arctan(24/7) - arctan(3/4)
= arctan(117/44) - x = y - x ... [Taking arctan(117/44) = y]

arctan (117/24) = arcsin(117/125) = arccos(44/125)
=> siny = 117/125 and cosy = 44/125

=> 7sinx + 24cosx = 7cosx + 20cos( y - x)
=> 7sinx + 17cosx = 20cosx cosy + 20sinx siny

=> (20siny - 7) sinx = (17 - 20cosy) cosx
=> tanx = (17 - 20cosy) / (20siny - 7)
=> tanx = [17 - 20 * (44/125)] / [20 * (117/125) - 7] = 249/293
=> sinx = 249/[5√(5914)] and cosx = 293/[5√(5914)]

=> Exact length of the side of the square
= 7sinx + 24cosx
= 7 * 249 / [5√(5914) + 24 * 293 / [5√(5914)]
= 1755 / √(5914)
≈ 22.82.

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Q.160. Puzzle

Question 160.
A room is 15m x 6m x 6m in length, width and height. A spider is on the middle of a 6 x 6 wall, 50cm down from the ceiling and sees a fly on the middle of the opposite 6 x 6 wall which is 50cm from the floor. What is the shortest distance the spider can travel to catch the fly?

Answer 160.
Just open up the room like the flaps of a box as shown in the figure below.

Four rectangles of the room of size 15m x 6m will appear side by side touching at lengths, the first being the
ceiling, next side-wall, then floor followed by side wall. Above the ceiling rectangle will be 6m x 6m end wall and below the floor rectangle will be the other 6m x 6m end wall.
 Plot the points on the end walls showing the initial position of the spider and the fly. Join them by a straight line. It forms the hypotenuse of a right triangle of sides 12m x 16m and its length is
 √[(12)^2 + (16)^2] = 20m which is the required shortest distance.

For better explanation, refer to the room in the form of the folded box below.

Room has been opened like a cardboard box and then a path of the spider is plotted. It should be remembered that every point on the hypotenuse is on some wall of the room and when spider moves on that line, it is moving on the wall and not flying. Now, refold the opened box and bring it back in the form of room. The line drawn is on different walls.

The spider moves a small distance on the end wall obliquely and goes to the ceiling, then again an oblique path coming to the side-wall, again obliquely moving it comes to the floor where also it moves obliquely to reach the end wall little away from where the fly is and reaches the fly moving at an agle on the end-wall. The hypotenuse drawn is the path which the spider follows and on refolding the walls back in the form of the room, the path is on the walls and not in air.

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