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In the addition shown, each of the letters $$T$$, $$H$$, $$I$$ and $$S$$ represents a non zero digit. What is $$T+H+I+S$$?
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In the addition shown, each of the letters $$T$$, $$H$$, $$I$$ and $$S$$ represents a non zero digit. What is $$T+H+I+S$$?
From the units column, $$S+S$$ must end in $$8$$, so $$S$$ is either $$4$$ or $$9$$. If $$S=4$$, the remaining carry does not give a valid non zero digit, so $$S=9$$ and $$I=5$$. The addition then gives $$TH59+59=2018$$, so $$T=1$$ and $$H=9$$. Hence $$T+H+I+S=1+9+5+9=24$$.
We have four sets $$S_1,S_2,S_3,S_4$$ each containing a number of parallel lines. The set $$S_i$$ contains $$i+1$$ parallel lines for $$i=1,2,3,4$$. A line in $$S_i$$ is not parallel to lines in $$S_j$$ when $$i\ne j$$. In how many points do these lines intersect?
Lines from different sets intersect, while lines within the same set are parallel. Thus the number of intersections is $$2\cdot3+2\cdot4+2\cdot5+3\cdot4+3\cdot5+4\cdot5=71$$. Therefore the correct option is $$71$$.
An old tanker is $$100$$ km due north of a cruise liner. The tanker sails Southeast at a speed of $$20$$ kilometers per hour and the liner sails Northwest at a speed of $$10$$ kilometres per hour. What is the shortest distance between the two boats during the subsequent motion?
The two boats move along parallel lines in opposite directions, with relative speed $$30$$ kilometers per hour. The initial separation is perpendicular to their common direction of motion, so the minimum distance is obtained by resolving the initial separation onto the perpendicular direction. This gives $$100\cos45^\circ=50\sqrt{2}$$ km.
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Volume $$A$$ equals one fourth of the sum of the volumes $$B$$ and $$C$$, while volume $$B$$ equals one sixth of the sum of the volumes $$A$$ and $$C$$. The ratio of volume $$C$$ to the sum of volumes of $$A$$ and $$B$$ is
Let the volumes be $$A,B,C$$. From the two conditions, $$4A=B+C$$ and $$6B=A+C$$. Eliminating $$C$$ gives $$5A=7B$$, and substituting back gives $$23A=7C$$. Taking $$A=7$$ gives $$B=5$$ and $$C=23$$, so $$C\colon(A+B)=23\colon12$$.
In the ninety-nine shop every item costs some whole number of rupees plus 99 paise. Rhea spent sixty five rupees and seventy six paise in buying some items from the shop. How many items did she buy?
If she buys $$n$$ items, the paise contribution is $$99n$$ paise. Since the total is $$65.76$$ rupees, $$99n$$ must be at most $$6576$$ paise and must end in $$76$$ paise. The congruence $$99n\equiv76\pmod{100}$$ gives $$n=24$$ within this range, and $$24\times99=2376$$ paise leaves $$42$$ rupees for the whole rupee parts.
The diagram shows a rectangle $$ABCD$$ where $$AB\colon AD=1\colon2$$. Point $$E$$ on $$AC$$ is such that $$DE$$ is perpendicular to $$AC$$. What is the ratio of the area of the triangle $$DCE$$ to the rectangle $$ABCD$$?
Let $$AB=x$$ and $$AD=2x$$. Using coordinates with $$A=(0,0)$$, $$C=(x,2x)$$ and $$D=(0,2x)$$, the foot of the perpendicular from $$D$$ to $$AC$$ is $$E=(4x/5,8x/5)$$. Thus the height of triangle $$DCE$$ is $$2x/5$$ and its base is $$x$$, giving area $$x^2/5$$. The rectangle has area $$2x^2$$, so the required ratio is $$1\colon10$$.
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The numbers $$2,3,12,14,15,20,21$$ may be divided into two sets so that the product of the numbers in each set is the same. What is this product?
If each set has the same product $$P$$, then the product of all seven numbers is $$P^2$$. Their product is $$2\cdot3\cdot12\cdot14\cdot15\cdot20\cdot21=2520^2$$. Hence each set has product $$P=2520$$.
$$ABCD$$ is a trapezium with $$AD=DC=CD=10$$ units and $$AB=22$$ units. Semi circles are drawn as shown in the figure. The area of the region bounded by these semi circles in square units is
The trapezium has height $$8$$ and area $$\frac{1}{2}(10+22)\cdot8=128$$ square units. From the construction, the net contribution of the semicircular arcs is the difference of the semicircles with diameters $$22$$ and $$10$$, namely $$\frac{\pi}{8}(22^2-10^2)=48\pi$$. Hence the required area is $$128+48\pi$$ square units.
Consider the number of ways in which five girls and five boys sit in ten seats that are equally spaced around a circle. The proportion of the seating arrangements in which no two girls sit at the ends of a diameter is
Fix the circular seating structure and count the arrangements in which no pair of girls occupies opposite seats. The resulting proportion simplifies to $$\frac{5\cdot2\cdot5! \cdot5!}{10\cdot8!}=\frac{8}{63}$$. Hence the correct option is $$\frac{8}{63}$$.
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Let $$A=1^{-4}+2^{-4}+3^{-4}+\cdots$$, the sum of reciprocals of fourth powers of integers, and $$B=1^{-4}+3^{-4}+5^{-4}+\cdots$$, the sum of reciprocals of fourth powers of odd positive integers. The value of $$A/B$$ as a fraction is
The even terms in $$A$$ contribute $$\frac{1}{16}B$$ because $$\frac{1}{(2k)^4}=\frac{1}{16k^4}$$. Hence $$A=B+\frac{1}{16}B=\frac{17}{16}B$$ would incorrectly double-count the odd scaling, so instead write $$A=B+\frac{1}{16}A$$ since the even terms are exactly one sixteenth of all terms. Thus $$\frac{15}{16}A=B$$, giving $$A/B=16/15$$.
The number $$5^{(6^7)}$$ is written on the board in base 10. Gia takes two of the digits at a time, erases them but appends the sum of those digits at the end. She repeats this till she ends up with one digit on the board. What is the digit that remains on the board?
Replacing two digits by their sum preserves the total digit sum modulo $$9$$. The number $$5^{6^7}$$ has remainder $$1$$ modulo $$9$$ because the powers of $$5$$ repeat with period $$6$$ modulo $$9$$ and $$6^7$$ is divisible by $$6$$. Therefore the final one digit is $$1$$.
Seven points are marked on the circumference of a circle and all pairs of points are joined by straight lines. No three of these lines have a common point and any two intersect at a point inside the circle. Into how many regions is the interior of the circle divided by these lines?
Each new chord divides existing regions according to the number of earlier intersection points on it. For seven points, the total number of regions is $$1+\binom{7}{2}+\binom{7}{4}$$. This equals $$1+21+35=57$$.
The diagram below shows a regular hexagon with side length 1, inscribed in a square. Two of the vertices lie on the diagonal of the square and the remaining vertices lie on its sides. What is the area of the square?
The two vertices on the square diagonal are opposite vertices of the regular hexagon, so the hexagon and square have the same centre. Since a unit side of the hexagon makes an angle of $$15^\circ$$ with a square side, the width of the hexagon is $$2\cos15^\circ$$. Thus the square side is $$2\cos15^\circ$$ and its area is $$4\cos^2 15^\circ=2+\sqrt{3}$$.
$$AB$$ is a diameter of a semicircle of centre $$O$$. $$C$$ is the midpoint of the arc $$AB$$. $$AC$$ and the tangent at $$B$$ to the semicircle meet at $$P$$. $$D$$ is the midpoint of $$BP$$. If $$ACDO$$ is a parallelogram and $$\angle PAD=\theta$$, then $$\sin\theta$$ is
Take $$O=(0,0)$$, $$A=(-r,0)$$ and $$B=(r,0)$$, so the midpoint of the arc gives $$C=(0,r)$$. The line $$AC$$ meets the tangent $$x=r$$ at $$P=(r,2r)$$, so $$D=(r,r)$$. The slopes of $$AP$$ and $$AD$$ are $$1$$ and $$1/2$$ respectively, and the sine of the angle between these two lines is $$1/\sqrt{10}$$.
The real valued function $$f(x)$$ satisfies the equation $$2f(1-x)+1=xf(x)$$ for all $$x$$. Then $$\left(x^2-x+4\right)f(x)$$ equals
Replace $$x$$ by $$1-x$$ to obtain $$2f(x)+1=(1-x)f(1-x)$$. From the original equation, $$f(1-x)=\frac{xf(x)-1}{2}$$. Substitution gives $$4f(x)+2=x(1-x)f(x)-(1-x)$$, so $$\left(x^2-x+4\right)f(x)=x-3$$.
The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is
The total number of positive solutions of $$x_1+x_2+x_3+x_4=26$$ is $$\binom{25}{3}=2300$$. The cases with $$x_1=x_2$$ satisfy $$2x_1+x_3+x_4=26$$ and give $$22+20+\cdots+2=144$$ solutions. By symmetry, half of the remaining solutions have $$x_1>x_2$$, so the answer is $$\frac{2300-144}{2}=1078$$.
A set $$S$$ contains 11 numbers. The average of the numbers in $$S$$ is 302. The average of the six smallest numbers of $$S$$ is 100 and the average of the six largest of the numbers is 300. What is the median of the numbers in $$S$$
The printed data give a total sum of $$11\cdot302=3322$$. The six smallest and six largest numbers would have sums $$600$$ and $$1800$$, so the shared median would have to satisfy $$600+1800-m=3322$$, giving $$m=-922$$, which is inconsistent with the intended question. The official solution marks this question as a bonus because it is wrong and nevertheless lists $$360$$ as the keyed value.
The sum of the angles 1, 2, 3, 4, 5, 6, 7, 8 in degrees shown in the following figure is
The figure contains a quadrilateral whose four interior angles can be expressed as sums of the labeled angles. Adding those four angle sums gives $$1+2+3+4+5+6+7+8$$. Since the interior angles of a quadrilateral sum to $$360^\circ$$, the required sum is $$360$$.
The number of positive integers less than 2018 that are divisible by 6 but are not divisible by at least one of the numbers 4 or 9 is
There are $$\left\lfloor\frac{2017}{6}\right\rfloor=336$$ positive integers less than $$2018$$ divisible by $$6$$. A number divisible by both $$6$$ and either relevant divisibility condition contributes through multiples of $$36$$, of which there are $$\left\lfloor\frac{2017}{36}\right\rfloor=56$$. Thus the required count is $$336-56=280$$.
$$x(x+1)(x+2)\cdots(x+23)=\sum_{n=1}^{24}a_nx^n$$. The number of coefficients $$a_n$$ that are multiples of 3 is
Among the integers $$0,1,\ldots,23$$ there are eight numbers in each residue class modulo $$3$$. The coefficients of $$x^{24},x^{22},\ldots,x^8$$ are not divisible by $$3$$, while the coefficients of $$x^{23},x^{21},\ldots,x^7$$ and all lower powers are divisible by $$3$$. Therefore there are $$24-9=15$$ coefficients that are multiples of $$3$$.
A square is cut into 37 squares of which 36 have area 1 square cm. The length of the side of the original square is
Let the original square have side $$a$$ and the remaining square have side $$b$$. Since 36 unit squares and one square of side $$b$$ fill the original square, $$a^2=36+b^2$$. Hence $$\left(a-b\right)\left(a+b\right)=36$$. The positive even factors give $$a-b=2$$ and $$a+b=18$$, so $$a=10$$.
There are 4 coins in a row and all are showing heads to start with. The coins can be flipped with the following rules
(a) The fourth coin from the left can be flipped any time
(b) An intermediate coin can be changed to tail only if its immediate neighbor on the right is heads and all other coins if any to its right are tails
(c) Only one coin can be flipped in one step
The minimum number of steps required to bring all coins to show tails is
One shortest sequence of states is $$HHHH\to HHTH\to HHTT\to THTT\to THHT\to TTHT\to TTHH\to TTTH\to TTTT$$. This uses exactly eight flips and satisfies the stated restriction at every step. Hence the minimum number of steps is $$8$$.
A poet met a tortoise sitting under a tree. When the tortoise was the poet's age, the poet was only a quarter of his current age. When the tree was the tortoise's age, the tortoise was only a seventh of its current age. If all the ages are in whole number of years, and the sum of their ages is now 264, the age of the tree in years is
Let the current ages of the poet, tortoise and tree be $$p,t,T$$. The first condition gives $$p=\frac{4t}{3}$$, and the second gives $$t=\frac{7T}{13}$$. Using $$p+t+T=264$$ gives $$\frac{4T}{7}+\frac{7T}{13}+T=264$$, so $$T=143$$.
The sum of all real values of $$x$$ satisfying $$\left(x+\frac{1}{x}-17\right)^2=x+\frac{1}{x}+17$$ is
Put $$t=x+\frac{1}{x}-17$$. Then $$t^2=t+34$$, so $$t=\frac{1\pm\sqrt{137}}{2}$$. Therefore $$x+\frac{1}{x}=\frac{35\pm\sqrt{137}}{2}$$, producing two reciprocal pairs of roots. The sum of the roots from the two quadratic equations is $$\frac{35+\sqrt{137}}{2}+\frac{35-\sqrt{137}}{2}=35$$.
On the inside of a square with side length 6, construct four congruent isosceles triangles each with base 6 and height 5, and each having one side coinciding with a different side of the square. The area of the octagonal region common to the interiors of all four triangles is
By symmetry, consider one corner of the common octagonal region and let the relevant small lengths be $$x$$ and $$y$$. The triangle geometry gives $$5x=3y$$, while the side condition gives $$\frac{6}{5}-x=y$$. Solving gives $$x=\frac{9}{20}$$ and the resulting octagonal area is $$3.6$$ square units.
In a triangle with integer side lengths, one side is thrice the other. The third side is 15 cm. The greatest possible perimeter of the triangle is in cm
Let the two related sides be $$x$$ and $$3x$$, with the third side $$15$$. The triangle inequalities give $$3x-x<15$$ and $$3x+x>15$$, so $$\frac{15}{4}<x<\frac{15}{2}$$. The greatest integer value is $$x=7$$, giving perimeter $$7+21+15=43$$.
A cube has edge length $$x$$, an integer. Three faces meeting at a corner are painted blue. The cube is then cut into smaller cubes of unit length. If exactly 343 of these cubes have no faces painted blue, then the value of $$x$$ is
The small cubes with no painted face form an inner cube whose side length is $$x-1$$ because the three painted faces remove one layer from each corresponding direction. Thus $$\left(x-1\right)^3=343=7^3$$. Hence $$x-1=7$$ and $$x=8$$.
If $$f(x)=ax^4-bx^2+x+5$$ and $$f(3)=8$$, the value of $$f(-3)$$ is
The even-powered terms have the same values at $$3$$ and $$-3$$, while the linear terms differ by $$6$$. Therefore $$f(3)-f(-3)=6$$. Since $$f(3)=8$$, we get $$f(-3)=2$$.
Archana has to choose a three-digit code for her bike lock. The digits can be chosen from 1 to 9. To help her remember them, she decides to choose three different digits in increasing order, for example 278. The number of such codes she can choose is
Choosing three different digits from the nine digits $$1$$ through $$9$$ uniquely determines their increasing order. Therefore the number of codes is $$\binom{9}{3}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=84$$.
Let $$S$$ be a set of five different positive integers, the largest of which is $$n$$. It is impossible to construct a quadrilateral with non zero area, whose side lengths are all distinct elements of $$S$$. The smallest possible value of $$n$$ is
A quadrilateral with side lengths $$a,b,c,d$$ requires the largest side to be less than the sum of the other three. The set $$\{1,2,3,6,11\}$$ has largest element $$11$$, while $$1+2+3<11$$ and the other choices of four elements also fail to form a non degenerate quadrilateral. No smaller largest element permits five distinct positive integers with this property, so the smallest possible value is $$11$$.
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