Lukujono

# Checking if it is one should be made more efficient, maybe by keeping track of it is one every time a change to V is made? only needs to be done on division
INCREASE U
REPEAT X TIMES ( # Sets Q to 1 if we should print anything after. if its zero, we dont do anything
    INCREASE V
	REPEAT O TIMES ( # Runs on second ( V > 1 ), sets Q = 1
		CLEAR O
		INCREASE Q
	)
	REPEAT U TIMES ( # Runs on first
		CLEAR Q
		INCREASE O
		CLEAR U
	)
)
PRINT X
CLEAR X
INCREASE E # Later should be replaced with 1000 or similar
INCREASE E
INCREASE E
INCREASE E
INCREASE E
INCREASE E
INCREASE E
INCREASE E
INCREASE E
INCREASE E
REPEAT E TIMES (
    INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
)
REPEAT E TIMES (
    INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
	INCREASE E
)
REPEAT E TIMES (
    REPEAT Q TIMES (
        CLEAR A
        INCREASE A
        CLEAR B
        CLEAR C
		CLEAR J
		INCREASE O
		CLEAR Q
        # Check for divisible by 2, output to X, Y AND J
        REPEAT V TIMES (
            REPEAT A TIMES (
				CLEAR A
				INCREASE B
				INCREASE X
				INCREASE J
				CLEAR Y
            )
            REPEAT C TIMES (
				CLEAR C
				INCREASE A
				CLEAR X
				INCREASE Y
				REPEAT O TIMES ( # Runs on second ( V > 1 ), sets Q = 1
					CLEAR O
					INCREASE Q # We increase Q here because 
				)
            )
            REPEAT B TIMES (
				CLEAR B
				INCREASE C
            )
        )
        REPEAT X TIMES ( # pariton, V = 3*V
            REPEAT V TIMES (
                INCREASE V
                INCREASE V
            )
            INCREASE V
        )
        REPEAT Y TIMES ( # parillinen, V = V/2
			CLEAR V
            REPEAT J TIMES (
			    INCREASE V
            )
        )
        PRINT V
    )
)

Test 1 (public)

Verdict: ACCEPTED