Find the smallest multiple of 15 such that each digit of the multiple is either 0 or 4

1 answer:

5 0

The answer is 4440. You can determine this by using your rules of divisibility. We know that is must end in a 0 or 5 to be divisible by 5 and that the sum of the digits must be divisible by 3 in order to divide by 3. One or two 4's together are not divisible by 3. Three 4's are, but it must end in 0 to be divisible by 5 as well.

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In the figure, AB║CD and ∠EIA is congruent to ∠GJB Complete the following statements to prove that ∠IKL is congruent to ∠DLH.

REY [17]

Here it is given that AB || CD

< EIA = <GJB

Now

∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)

∠EIA ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)

∠IKL + ∠IKC 180° and ∠DLH + ∠JLD =180° (Linear Pair Theorem)

m∠IKL + m∠IKC = 180° ....(1)

But ∠IKC ≅ ∠JLD

m∠IKC = m∠JLD (SUBTRACTION PROPERTY OF CONGRUENCY)