Counting 1

randRange( 20, 50 ) randRange( 20, 40 ) START + ANSWER - 1 spin( "{do,did,done,problem|read,read,read,{page|paper|article}}" ).split( "," ) plural( PROBLEM )

person( 1 ) {had to DO|was assigned} PROBLEMS START through END for homework {{last {week|night}}|tonight}.

If person( 1 ) DID all of the PROBLEMS he( 1 ) was assigned, how many PROBLEMS did he( 1 ) DO?

ANSWER PROBLEMS

Instead of counting PROBLEMS START through END, we can subtract START - 1 from each number so we can count the PROBLEMS starting from 1.

\qquad \array{ \hspace{40px} & \hspace{40px} & \hspace{40px} & \hspace{40px} & \\ \color{GREEN}{START} & \color{GREEN}{START + 1} & \color{GREEN}{START + 2} & & \color{GREEN}{END} \\ \color{BLUE}{\downarrow} & \color{BLUE}{\downarrow} & \color{BLUE}{\downarrow} & & \color{BLUE}{\downarrow} \\ \color{BLUE}{\llap{-}START-1} & \color{BLUE}{\llap{-}START-1} & \color{BLUE}{\llap{-}START-1} & \cdots\hphantom{-} & \color{BLUE}{\llap{-}START-1} \\ \color{BLUE}{\downarrow} & \color{BLUE}{\downarrow} & \color{BLUE}{\downarrow} & & \color{BLUE}{\downarrow} \\ \color{PINK}{1} & \color{PINK}{2} & \color{PINK}{3} & & \color{PINK}{ANSWER} \\ }

Now instead of thinking about PROBLEMS START through END, we can think about how many PROBLEMS person( 1 ) would have DONE if he( 1 ) were assigned PROBLEMS 1 though ANSWER.

When we think about it this way, we can see that person( 1 ) would have DONE ANSWER PROBLEMS. Hopefully it also makes sense why we can't just subtract the first problem number from the last problem number, since if person( 1 ) were assigned PROBLEMS 1 through ANSWER, he( 1 ) was assigned ANSWER PROBLEMS, not ANSWER-1.

person( 1 ) DID ANSWER PROBLEMS.

randRange( 5, 15 )

A baker has a whole {baguette|loaf of bread}.

How many cuts must {he|she} make to have exactly NUM even slices?

NUM -1 cuts

init({ range: [ [ -0.5, 17.5 ], [ -4, -0.5 ] ], scale: [ 27, 50 ] }); graph.loaf = raphael.set(); graph.labels = []; graph.drawLoaf = function( slices ) { var width = 10 + ( slices - 1 ) * 0.5; graph.loaf.remove(); jQuery.each( graph.labels, function() { this.remove(); }); style({ stroke: "black", fill: ORANGE, opacity: 0.3 }, function() { graph.loaf.push( arc([ 0.6, -2 ], 1, 90, 270, false ) ); graph.loaf.push( arc([ width - 0.6, -2 ], 1, 270, 90, false ) ); for ( var slice = 0; slice < slices; slice++ ) { if ( slice === 0 ) { graph.loaf.push( path([ [ 0.6, -1 ], [ 10 * ( slice + 1 ) / slices + slice * 0.5, -1 ], [ 10 * ( slice + 1 ) / slices + slice * 0.5, -3 ], [ 0.6, -3 ] ]) ); } else if ( slice === slices - 1 ) { graph.loaf.push( path([ [ 10 * ( slice + 1 ) / slices + slice * 0.5 - 0.6, -3 ], [ 10 * slice / slices + slice * 0.5, -3 ], [ 10 * slice / slices + slice * 0.5, -1 ], [ 10 * ( slice + 1 ) / slices + slice * 0.5 - 0.6, -1 ] ]) ); } else { graph.loaf.push( path([ [ 10 * slice / slices + slice * 0.5, -1 ], [ 10 * ( slice + 1 ) / slices + slice * 0.5, -1 ], [ 10 * ( slice + 1 ) / slices + slice * 0.5, -3 ], [ 10 * slice / slices + slice * 0.5, -3 ], [ 10 * slice / slices + slice * 0.5, -1 ] ]) ); } graph.labels.push( label( [ 10 * ( slice + 0.5 ) / slices + slice * 0.5, -2 ], slice+1, "", false ) .css({ fontWeight: "bold", opacity: 1 }) ); if ( slice !== 0 ) { graph.labels.push( label( [ 10 * slice / slices + slice * 0.5 - 0.2, -3 ], slice, "below", false ) .addClass( "hint_blue" ).css({ fontWeight: "bold", opacity: 1 }) ); } } }); return loaf; }; graph.drawLoaf( 2 );

One cut will make two slices.

graph.drawLoaf( 3 );

Two cuts will make three slices, and so on.

graph.drawLoaf( NUM );

Therefore, we need NUM - 1 cuts to make NUM slices.

randRange( 10, 20 )

person( 1 ) is building a straight fence, with posts one meter apart.

If the fence is LENGTH meters long, how many fence posts does he( 1 ) need?

LENGTH + 1 fence posts

init({ range: [ [ -1, LENGTH + 1 ], [ -2, 3.5 ] ], scale: [ 475 / ( LENGTH + 2 ), 20 ] }); style({ stroke: null, fill: BLUE, opacity: 0.3 }, function() { path([ [ 0, 0 ], [ 0, 1.8 ], [ 1, 1.8 ], [ 1, 0 ], [ 0, 0 ] ]); }); style({ stroke: GREEN, strokeWidth: 3 }, function() { line( [ 0, 0 ], [ 0, 2 ] ); line( [ 1, 0 ], [ 1, 2 ] ); }); label( [ 0, 1.8 ], "1", "above", false ).css({ color: GREEN }); label( [ 1, 1.8 ], "2", "above", false ).css({ color: GREEN }); style({ stroke: BLUE, strokeWidth: 2 }, function() { graph.brace = path([ [ 0, -0.2 ], [ 0, -0.4 ], [ 0.5, -0.4 ], [ 0.5, -0.6 ], [ 0.5, -0.4 ], [ 1, -0.4 ], [ 1, -0.2 ] ]); }); graph.length = label( [ 0.5, -0.4 ], "1 meter", "below", false ).css({ color: BLUE });

If the fence is one meter long, he( 1 ) needs two posts (one for each end).

style({ stroke: null, fill: BLUE, opacity: 0.3 }, function() { path([ [ 1, 0 ], [ 1, 1.8 ], [ 2, 1.8 ], [ 2, 0 ], [ 1, 0 ] ]); }); style({ stroke: GREEN, strokeWidth: 3 }, function() { line( [ 1, 0 ], [ 1, 2 ] ); line( [ 2, 0 ], [ 2, 2 ] ); }); label( [ 2, 1.8 ], "3", "above", false ).css({ color: GREEN }); graph.brace.remove(); graph.length.remove(); style({ stroke: BLUE, strokeWidth: 2 }, function() { graph.brace = path([ [ 0, -0.2 ], [ 0, -0.4 ], [ 1, -0.4 ], [ 1, -0.6 ], [ 1, -0.4 ], [ 2, -0.4 ], [ 2, -0.2 ] ]); }); graph.length = label( [ 1, -0.4 ], "2 meters", "below", false ).css({ color: BLUE });

If the fence is two meters long, then he( 1 ) needs three posts, and so on.

for ( var x = 2; x < LENGTH; x++ ) { style({ stroke: null, fill: BLUE, opacity: 0.3 }, function() { path([ [ x, 0 ], [ x, 1.8 ], [ x + 1, 1.8 ], [ x + 1, 0 ], [ x, 0 ] ]); }); style({ stroke: GREEN, strokeWidth: 3 }, function() { line( [ x, 0 ], [ x, 2 ] ); line( [ x + 1, 0 ], [ x + 1, 2 ] ); }); label( [ x + 1, 1.8 ], x + 2, "above", false ).css({ color: GREEN }); } graph.brace.remove(); graph.length.remove(); style({ stroke: BLUE, strokeWidth: 2 }, function() { graph.brace = path([ [ 0, -0.2 ], [ 0, -0.4 ], [ LENGTH/2, -0.4 ], [ LENGTH/2, -0.6 ], [ LENGTH/2, -0.4 ], [ LENGTH, -0.4 ], [ LENGTH, -0.2 ] ]); }); graph.length = label( [ LENGTH / 2, -0.4 ], LENGTH + " meters", "below", false ).css({ color: BLUE });

Therefore, for his( 1 ) LENGTH meter fence, person( 1 ) needs LENGTH + 1 posts.