randRangeWeighted(0, 9, 0, 0.5) [randRange(0, 9), randRange(0, 9), randRange(0, 9)] [rand(3) ? randRange(0, 9) : UNIT].concat(shuffle(NUMS)) [rand(3) ? randRange(0, 9) : UNIT].concat(shuffle(NUMS)) 1000 * DIGITS1[3] + DIGITS1[2] * 100 + DIGITS1[1] * 10 + DIGITS1[0] 1000 * DIGITS2[3] + DIGITS2[2] * 100 + DIGITS2[1] * 10 + DIGITS2[0] D1 / 1000 D2 / 1000 (function() { if (N1 === N2) { return "="; } else { return N1 < N2 ? "<" : ">"; } })()

Fill in the blank.

\large{N1} ____ \large{N2}

SOLUTION

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[randRange(0, 9), randRange(0, 9)] (function() { var arr = shuffle(NUMS); arr.splice(rand(2), 0, 0); arr.push(0); return arr; })() (function() { var arr = shuffle(NUMS); arr.splice(rand(2), 0, 0); arr.push(0); return arr; })() DIGITS1[3] === 0 && rand(2) ? "0" : "" DIGITS1[3] === 0 && rand(2) ? "0" : "" 1000 * DIGITS1[3] + DIGITS1[2] * 100 + DIGITS1[1] * 10 + DIGITS1[0] 1000 * DIGITS2[3] + DIGITS2[2] * 100 + DIGITS2[1] * 10 + DIGITS2[0]
D1 / 1000 D2 / 1000 (function() { if (N1 === N2) { return "="; } else { return N1 < N2 ? "<" : ">"; } })()

Fill in the blank.

\large{N1TRAILING1} ____ \large{N2TRAILING2}

SOLUTION

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\large{<} means "less than".
\large{>} means "greater than".

\large{N1 = DIGITS1[3] + \dfrac{DIGITS1[2]}{10} + \dfrac{DIGITS1[1]}{100} + \dfrac{DIGITS1[0]}{1000}}

\large{\phantom{N1} = \dfrac{1000 * DIGITS1[3]}{1000} + \dfrac{100 * DIGITS1[2]}{1000} + \dfrac{10 * DIGITS1[1]}{1000} + \dfrac{DIGITS1[0]}{1000}}

\large{\phantom{N1} = \dfrac{1000 * DIGITS1[3] + 100 * DIGITS1[2] + 10 * DIGITS1[1] + DIGITS1[0]}{1000}}

\large{\phantom{N1} = \dfrac{D1}{1000}}

\large{N2 = DIGITS2[3] + \dfrac{DIGITS2[2]}{10} + \dfrac{DIGITS2[1]}{100} + \dfrac{DIGITS2[0]}{1000}}

\large{\phantom{N2} = \dfrac{1000 * DIGITS2[3]}{1000} + \dfrac{100 * DIGITS2[2]}{1000} + \dfrac{10 * DIGITS2[1]}{1000} + \dfrac{DIGITS2[0]}{1000}}

\large{\phantom{N2} = \dfrac{1000 * DIGITS2[3] + 100 * DIGITS2[2] + 10 * DIGITS2[1] + DIGITS2[0]}{1000}}

\large{\phantom{N2} = \dfrac{D2}{1000}}

\large{D1} thousandths is more than \large{D2} thousandths.

\large{D1} thousandths is less than \large{D2} thousandths.

Therefore \large{N1 > N2}.

Therefore \large{N1 < N2}.